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SQL interview questions and answers

What is SQL and where does it come from?
Structured Query Language (SQL) is a language that provides an interface to relational database systems.
SQL was developed by IBM in the 1970s for use in System R, and is a de facto standard, as well as an ISO
and ANSI standard. SQL is often pronounced SEQUEL.
In common usage SQL also encompasses DML (Data Manipulation Language), for INSERTs, UPDATEs,
DELETEs and DDL (Data Definition Language), used for creating and modifying tables and other database
structures.
The development of SQL is governed by standards. A major revision to the SQL standard was completed in
1992, called SQL2. SQL3 support object extensions and are (partially?) implemented in Oracle8 and 9.

What are the difference between DDL, DML and DCL commands?
DDL is Data Definition Language statements. Some examples:
• CREATE - to create objects in the database
• ALTER - alters the structure of the database
• DROP - delete objects from the database
• TRUNCATE - remove all records from a table, including all spaces allocated for the records are
removed
• COMMENT - add comments to the data dictionary
• GRANT - gives user's access privileges to database
• REVOKE - withdraw access privileges given with the GRANT command
DML is Data Manipulation Language statements. Some examples:
• SELECT - retrieve data from the a database
• INSERT - insert data into a table
• UPDATE - updates existing data within a table
• DELETE - deletes all records from a table, the space for the records remain
• CALL - call a PL/SQL or Java subprogram
• EXPLAIN PLAN - explain access path to data
• LOCK TABLE - control concurrency
DCL is Data Control Language statements. Some examples:
• COMMIT - save work done
• SAVEPOINT - identify a point in a transaction to which you can later roll back
• ROLLBACK - restore database to original since the last COMMIT
• SET TRANSACTION - Change transaction options like what rollback segment to use

How does one escape special characters when building SQL queries?
The LIKE keyword allows for string searches. The '_' wild card character is used to match exactly one
character, '%' is used to match zero or more occurrences of any characters. These characters can be escaped
in SQL. Example:
SELECT name FROM emp WHERE id LIKE '%\_%' ESCAPE '\';
Use two quotes for every one displayed. Example:
SELECT 'Franks''s Oracle site' FROM DUAL;
SELECT 'A ''quoted'' word.' FROM DUAL;
SELECT 'A ''''double quoted'''' word.' FROM DUAL;

How does one eliminate duplicates rows from a table?
Choose one of the following queries to identify or remove duplicate rows from a table leaving only unique
records in the table:
Method 1:
SQL> DELETE FROM table_name A WHERE ROWID > (
2 SELECT min(rowid) FROM table_name B
3 WHERE A.key_values = B.key_values);
Method 2:
SQL> create table table_name2 as select distinct * from table_name1;
SQL> drop table_name1;
SQL> rename table_name2 to table_name1;
SQL> -- Remember to recreate all indexes, constraints, triggers, etc
on table...
Method 3: (thanks to Dennis Gurnick)
SQL> delete from my_table t1
SQL> where exists (select 'x' from my_table t2
SQL> where t2.key_value1 = t1.key_value1
SQL> and t2.key_value2 = t1.key_value2
SQL> and t2.rowid > t1.rowid);
Note: One can eliminate N^2 unnecessary operations by creating an index on the joined fields in the inner
loop (no need to loop through the entire table on each pass by a record). This will speed-up the deletion
process.
Note 2: If you are comparing NOT-NULL columns, use the NVL function. Remember that NULL is not
equal to NULL. This should not be a problem as all key columns should be NOT NULL by definition.

How does one generate primary key values for a table?
Create your table with a NOT NULL column (say SEQNO). This column can now be populated with
unique values:
SQL> UPDATE table_name SET seqno = ROWNUM;
or use a sequences generator:
SQL> CREATE SEQUENCE sequence_name START WITH 1 INCREMENT BY 1;
SQL> UPDATE table_name SET seqno = sequence_name.NEXTVAL;
Finally, create a unique index on this column.

How does one get the time difference between two date columns?
Look at this example query:
select floor(((date1-date2)*24*60*60)/3600)
|| ' HOURS ' ||
floor((((date1-date2)*24*60*60) -
floor(((date1-date2)*24*60*60)/3600)*3600)/60)
|| ' MINUTES ' ||
round((((date1-date2)*24*60*60) -
floor(((date1-date2)*24*60*60)/3600)*3600 -
(floor((((date1-date2)*24*60*60) -
floor(((date1-date2)*24*60*60)/3600)*3600)/60)*60)))
|| ' SECS ' time_difference
from ...
If you don't want to go through the floor and ceiling math, try this method (contributed by Erik Wile):
select to_char(to_date('00:00:00','HH24:MI:SS') +
(date1 - date2), 'HH24:MI:SS') time_difference
from ...
Note that this query only uses the time portion of the date and ignores the date itself. It will thus never
return a value bigger than 23:59:59.

How does one add a day/hour/minute/second to a date value?
The SYSDATE pseudo-column shows the current system date and time. Adding 1 to SYSDATE will
advance the date by 1 day. Use fractions to add hours, minutes or seconds to the date. Look at these
examples:
SQL> select sysdate, sysdate+1/24, sysdate +1/1440, sysdate +
1/86400 from dual;
SYSDATE SYSDATE+1/24 SYSDATE+1/1440
SYSDATE+1/86400
-------------------- -------------------- --------------------
--------------------
03-Jul-2002 08:32:12 03-Jul-2002 09:32:12 03-Jul-2002 08:33:12
03-Jul-2002 08:32:13
The following format is frequently used with Oracle Replication:
select sysdate NOW, sysdate+30/(24*60*60) NOW_PLUS_30_SECS from
dual;
NOW NOW_PLUS_30_SECS
-------------------- --------------------
03-JUL-2002 16:47:23 03-JUL-2002 16:47:53

How does one count different data values in a column?
Use this simple query to count the number of data values in a column:
select my_table_column, count(*)
from my_table
group by my_table_column;
A more sophisticated example...
select dept, sum( decode(sex,'M',1,0)) MALE,
sum( decode(sex,'F',1,0)) FEMALE,
count(decode(sex,'M',1,'F',1)) TOTAL
from my_emp_table
group by dept;

How does one count/sum RANGES of data values in a column?
A value x will be between values y and z if GREATEST(x, y) = LEAST(x, z). Look at this example:
select f2,
sum(decode(greatest(f1,59), least(f1,100), 1, 0)) "Range
60-100",
sum(decode(greatest(f1,30), least(f1, 59), 1, 0)) "Range
30-59",
sum(decode(greatest(f1, 0), least(f1, 29), 1, 0)) "Range
00-29"
from my_table
group by f2;
For equal size ranges it might be easier to calculate it with DECODE(TRUNC(value/range), 0, rate_0, 1,
rate_1, ...). Eg.
select ename "Name", sal "Salary",
decode( trunc(f2/1000, 0), 0, 0.0,
1, 0.1,
2, 0.2,
3, 0.31) "Tax rate"
from my_table;

Can one retrieve only the Nth row from a table?
Shaik Khaleel provided this solution to select the Nth row from a table:
SELECT * FROM (
SELECT ENAME,ROWNUM RN FROM EMP WHERE ROWNUM < 101 )
WHERE RN = 100;
Note: Note: In this first it select only one more than the required row, then it selects the required one. Its far
better than using MINUS operation.
Ravi Pachalla provided this solution:
SELECT f1 FROM t1
WHERE rowid = (
SELECT rowid FROM t1
WHERE rownum <= 10
MINUS
SELECT rowid FROM t1
WHERE rownum < 10);
Alternatively...
SELECT * FROM emp WHERE rownum=1 AND rowid NOT IN
(SELECT rowid FROM emp WHERE rownum < 10);
Please note, there is no explicit row order in a relational database. However, this query is quite fun and may
even help in the odd situation.

Can one retrieve only rows X to Y from a table?
Shaik Khaleel provided this solution to the problem:
SELECT * FROM (
SELECT ENAME,ROWNUM RN FROM EMP WHERE ROWNUM < 101
) WHERE RN between 91 and 100 ;
Note: the 101 is just one greater than the maximum row of the required rows (means x= 90, y=100, so the
inner values is y+1).
Another solution is to use the MINUS operation. For example, to display rows 5 to 7, construct a query like
this:
SELECT *
FROM tableX
WHERE rowid in (
SELECT rowid FROM tableX
WHERE rownum <= 7
MINUS
SELECT rowid FROM tableX
WHERE rownum < 5);
Please note, there is no explicit row order in a relational database. However, this query is quite fun and may
even help in the odd situation.

How does one select EVERY Nth row from a table?
One can easily select all even, odd, or Nth rows from a table using SQL queries like this:
Method 1: Using a subquery
SELECT *
FROM emp
WHERE (ROWID,0) IN (SELECT ROWID, MOD(ROWNUM,4)
FROM emp);
Method 2: Use dynamic views (available from Oracle7.2):
SELECT *
FROM ( SELECT rownum rn, empno, ename
FROM emp
) temp
WHERE MOD(temp.ROWNUM,4) = 0;
Please note, there is no explicit row order in a relational database. However, these queries are quite fun and
may even help in the odd situation.

How does one select the TOP N rows from a table?
Form Oracle8i one can have an inner-query with an ORDER BY clause. Look at this example:
SELECT *
FROM (SELECT * FROM my_table ORDER BY col_name_1 DESC)
WHERE ROWNUM < 10;
Use this workaround with prior releases:
SELECT *
FROM my_table a
WHERE 10 >= (SELECT COUNT(DISTINCT maxcol)
FROM my_table b
WHERE b.maxcol >= a.maxcol)
ORDER BY maxcol DESC;

How does one code a tree-structured query?
Tree-structured queries are definitely non-relational (enough to kill Codd and make him roll in his grave).
Also, this feature is not often found in other database offerings.
The SCOTT/TIGER database schema contains a table EMP with a self-referencing relation (EMPNO and
MGR columns). This table is perfect for tesing and demonstrating tree-structured queries as the MGR
column contains the employee number of the "current" employee's boss.
The LEVEL pseudo-column is an indication of how deep in the tree one is. Oracle can handle queries with
a depth of up to 255 levels. Look at this example:
select LEVEL, EMPNO, ENAME, MGR
from EMP
connect by prior EMPNO = MGR
start with MGR is NULL;
One can produce an indented report by using the level number to substring or lpad() a series of spaces, and
concatenate that to the string. Look at this example:
select lpad(' ', LEVEL * 2) || ENAME ........
One uses the "start with" clause to specify the start of the tree. More than one record can match the starting
condition. One disadvantage of having a "connect by prior" clause is that you cannot perform a join to other
tables. The "connect by prior" clause is rarely implemented in the other database offerings. Trying to do
this programmatically is difficult as one has to do the top level query first, then, for each of the records
open a cursor to look for child nodes.
One way of working around this is to use PL/SQL, open the driving cursor with the "connect by prior"
statement, and the select matching records from other tables on a row-by-row basis, inserting the results
into a temporary table for later retrieval.

How does one code a matrix report in SQL?
Look at this example query with sample output:
SELECT *
FROM (SELECT job,
sum(decode(deptno,10,sal)) DEPT10,
sum(decode(deptno,20,sal)) DEPT20,
sum(decode(deptno,30,sal)) DEPT30,
sum(decode(deptno,40,sal)) DEPT40
FROM scott.emp
GROUP BY job)
ORDER BY 1;
JOB DEPT10 DEPT20 DEPT30 DEPT40
--------- ---------- ---------- ---------- ----------
ANALYST 6000
CLERK 1300 1900 950
MANAGER 2450 2975 2850
PRESIDENT 5000
SALESMAN 5600

How does one implement IF-THEN-ELSE in a select statement?
The Oracle decode function acts like a procedural statement inside an SQL statement to return different
values or columns based on the values of other columns in the select statement.
Some examples:
select decode(sex, 'M', 'Male', 'F', 'Female', 'Unknown')
from employees;
select a, b, decode( abs(a-b), a-b, 'a > b',
0, 'a = b',
'a < b') from tableX;
select decode( GREATEST(A,B), A, 'A is greater OR EQUAL than
B', 'B is greater than A')...
select decode( GREATEST(A,B),
A, decode(A, B, 'A NOT GREATER THAN B', 'A GREATER
THAN B'),
'A NOT GREATER THAN B')...
Note: The decode function is not ANSI SQL and is rarely implemented in other RDBMS offerings. It is
one of the good things about Oracle, but use it sparingly if portability is required.
From Oracle 8i one can also use CASE statements in SQL. Look at this example:
SELECT ename, CASE WHEN sal>1000 THEN 'Over paid' ELSE 'Under
paid' END
FROM emp;
How can one dump/ examine the exact content of a database column?

SELECT DUMP(col1)
FROM tab1
WHERE cond1 = val1;
DUMP(COL1)
-------------------------------------
Typ=96 Len=4: 65,66,67,32
For this example the type is 96, indicating CHAR, and the last byte in the column is 32, which is the ASCII
code for a space. This tells us that this column is blank-padded.

Can one drop a column from a table?
From Oracle8i one can DROP a column from a table. Look at this sample script, demonstrating the ALTER
TABLE table_name DROP COLUMN column_name; command.
Other workarounds:
1. SQL> update t1 set column_to_drop = NULL;
SQL> rename t1 to t1_base;
SQL> create view t1 as select <specific columns> from t1_base;
2. SQL> create table t2 as select <specific columns> from t1;
SQL> drop table t1;
SQL> rename t2 to t1;

Can one rename a column in a table?
No, this is listed as Enhancement Request 163519. Some workarounds:
1. -- Use a view with correct column names...
rename t1 to t1_base;
create view t1 <column list with new name> as select * from t1_base;
2. -- Recreate the table with correct column names...
create table t2 <column list with new name> as select * from t1;
drop table t1;
rename t2 to t1;
3. -- Add a column with a new name and drop an old column...
alter table t1 add ( newcolame datatype );
update t1 set newcolname=oldcolname;
alter table t1 drop column oldcolname;

How can I change my Oracle password?
Issue the following SQL command: ALTER USER <username> IDENTIFIED BY
<new_password>
/
From Oracle8 you can just type "password" from SQL*Plus, or if you need to change another user's
password, type "password user_name".

How does one find the next value of a sequence?
Perform an "ALTER SEQUENCE ... NOCACHE" to unload the unused cached sequence numbers from the
Oracle library cache. This way, no cached numbers will be lost. If you then select from the
USER_SEQUENCES dictionary view, you will see the correct high water mark value that would be
returned for the next NEXTVALL call. Afterwards, perform an "ALTER SEQUENCE ... CACHE" to
restore caching.
You can use the above technique to prevent sequence number loss before a SHUTDOWN ABORT, or any
other operation that would cause gaps in sequence values.
Workaround for snapshots on tables with LONG columns
You can use the SQL*Plus COPY command instead of snapshots if you need to copy LONG and LONG
RAW variables from one location to another. Eg:
COPY TO SCOTT/TIGER@REMOTE -
CREATE IMAGE_TABLE USING -
SELECT IMAGE_NO, IMAGE -
FROM IMAGES;
Note: If you run Oracle8, convert your LONGs to LOBs, as it can be replicated.




What is SQL and where does it come from?
Structured Query Language (SQL) is a language that provides an interface to relational database systems. The proper pronunciation of SQL, and the preferred pronunciation within Oracle Corp, is "sequel" and not "ess cue ell".
SQL was developed by IBM in the 1970s for use in System R, and is a de facto standard, as well as an ISO and ANSI standard.
In common usage SQL also encompasses DML (Data Manipulation Language), for INSERTs, UPDATEs, DELETEs and DDL (Data Definition Language), used for creating and modifying tables and other database structures.
The development of SQL is governed by standards. A major revision to the SQL standard was completed in 1992, called SQL2. SQL3 support object extensions and are (partially?) implemented in Oracle8 and 9i.
Example SQL statements:


CREATE TABLE table1 (column1 NUMBER, column2 VARCHAR2(30));
INSERT INTO table1 VALUES (1, 'XYZ');
SELECT * FROM table1 WHERE column2 = 'XYZ';

What are the difference between DDL, DML and DCL commands?

DDL - Data Definition Language: statements used to define the database structure or schema. Some examples:
- CREATE - to create objects in the database
- ALTER - alters the structure of the database
- DROP - delete objects from the database
- TRUNCATE - remove all records from a table, including all spaces allocated for the records are removed
- COMMENT - add comments to the data dictionary
- RENAME - rename an object
DML - Data Manipulation Language: statements used for managing data within schema objects. Some examples:
- SELECT - retrieve data from the a database
- INSERT - insert data into a table
- UPDATE - updates existing data within a table
- DELETE - deletes all records from a table, the space for the records remain
- MERGE - UPSERT operation (insert or update)
- CALL - call a PL/SQL or Java subprogram
- EXPLAIN PLAN - explain access path to the data
- LOCK TABLE - controls concurrency
DCL - Data Control Language. Some examples:
- GRANT - gives user's access privileges to database
- REVOKE - withdraw access privileges given with the GRANT command
TCL - Transaction Control: statements used to manage the changes made by DML statements. It allows statements to be grouped together into logical transactions.
- COMMIT - save work done
- SAVEPOINT - identify a point in a transaction to which you can later roll back
- ROLLBACK - undo the modification I made since the last COMMIT
- SET TRANSACTION - Change transaction options like isolation level and what rollback segment to use
- SET ROLE - set the current active roles
DML are not auto-commit. i.e. you can roll-back the operations, but DDL are auto-commit


Difference between TRUNCATE, DELETE and DROP commands?

The DELETE command is used to remove some or all rows from a table. A WHERE clause can be used to only remove some rows. If no WHERE condition is specified, all rows will be removed. After performing a DELETE operation you need to COMMIT or ROLLBACK the transaction to make the change permanent or to undo it. Note that this operation will cause all DELETE triggers on the table to fire.


SQL> SELECT COUNT(*) FROM emp;
  COUNT(*)
----------
        14

SQL> DELETE FROM emp WHERE job = 'CLERK';
4 rows deleted.

SQL> COMMIT;
Commit complete.

SQL> SELECT COUNT(*) FROM emp;
  COUNT(*)
----------
        10

TRUNCATE removes all rows from a table. The operation cannot be rolled back and no triggers will be fired. As such, TRUNCATE is faster and doesn't use as much undo space as a DELETE.


SQL> TRUNCATE TABLE emp;
Table truncated.

SQL> SELECT COUNT(*) FROM emp;

  COUNT(*)
----------
         0

The DROP command removes a table from the database. All the tables' rows, indexes and privileges will also be removed. No DML triggers will be fired. The operation cannot be rolled back.


SQL> DROP TABLE emp;
Table dropped.

SQL> SELECT * FROM emp;
SELECT * FROM emp
              *
ERROR at line 1:
ORA-00942: table or view does not exist

DROP and TRUNCATE are DDL commands, whereas DELETE is a DML command. Therefore DELETE operations can be rolled back (undone), while DROP and TRUNCATE operations cannot be rolled back.
From Oracle 10g a table can be "undropped". Example:


SQL> FLASHBACK TABLE emp TO BEFORE DROP;
Flashback complete.

PS: DELETE will not free up used space within a table. This means that repeated DELETE commands will severely fragment the table and queries will have to navigate this "free space" in order to retrieve rows.


How does one escape special characters when writing SQL queries?

Escape quotes
Use two quotes for every one displayed. Examples:


SQL> SELECT 'Frank''s Oracle site' AS text FROM DUAL;
 TEXT
 --------------------
 Franks's Oracle site

 SQL> SELECT 'A ''quoted'' word.' AS text FROM DUAL;
 TEXT
 ----------------
 A 'quoted' word.

 SQL> SELECT 'A ''''double quoted'''' word.' AS text FROM DUAL;
 TEXT
 -------------------------
 A ''double quoted'' word.

Use Q expression:


SQL> SELECT q'[Frank's Oracle site]' AS text FROM DUAL;
 TEXT
 -------------------
 Frank's Oracle site

 SQL> SELECT q'[A 'quoted' word.]' AS text FROM DUAL;
 TEXT
 ----------------
 A 'quoted' word.

 SQL> SELECT q'[A ''double quoted'' word.]' AS text FROM DUAL;
 TEXT
 -------------------------
 A ''double quoted'' word.

Escape wildcard characters
The LIKE keyword allows for string searches. The '_' wild card character is used to match exactly one character, while '%' is used to match zero or more occurrences of any characters. These characters can be escaped in SQL. Examples:


SELECT name FROM emp
 WHERE id LIKE '%/_%' ESCAPE '/';
SELECT name FROM emp
 WHERE id LIKE '%\%%' ESCAPE '\';

Escape ampersand (&) characters in SQL*Plus
When using SQL*Plus, the DEFINE setting can be changed to allow &'s (ampersands) to be used in text:


SET DEFINE ~
SELECT 'Laurel & Hardy' FROM dual;

Other methods:
Define an escape character:


SET ESCAPE '\'
SELECT '\&abc' FROM dual;

Don't scan for substitution variables:


SET SCAN OFF
SELECT '&ABC' x FROM dual;

Another way to escape the & would be to use concatenation, which would not require any SET commands -


SELECT 'Laurel ' || '&' || ' Hardy' FROM dual;

Use the 10g Quoting mechanism:


Syntax
 q'[QUOTE_CHAR]Text[QUOTE_CHAR]'
 Make sure that the QUOTE_CHAR followed by an ' doesn't exist in the text.
SELECT q'{This is Orafaq's 'quoted' text field}' FROM DUAL;

Can one select a random collection of rows from a table?

The following methods can be used to select a random collection of rows from a table:
The SAMPLE Clause
From Oracle 8i, the easiest way to randomly select rows from a table is to use the SAMPLE clause with a SELECT statement. Examples:


SELECT * FROM emp SAMPLE(10);

In the above example, Oracle is instructed to randomly return 10% of the rows in the table.


SELECT * FROM emp SAMPLE(5) BLOCKS;

This example will sample 5% of all formatted database blocks instead of rows.
This clause only works for single table queries on local tables. If you include the SAMPLE clause within a multi-table or remote query, you will get a parse error or "ORA-30561: SAMPLE option not allowed in statement with multiple table references". One way around this is to create an inline view on the driving table of the query with the SAMPLE clause. Example:


SELECT t1.dept, t2.emp
  FROM (SELECT * FROM dept SAMPLE(5)) t1,
       emp t2
 WHERE t1.dep_id = t2.dep_id;

If you examine the execution plan of a "Sample Table Scan", you should see a step like this:


TABLE ACCESS (SAMPLE) OF 'EMP' (TABLE)

ORDER BY dbms_random.value()
This method orders the data by a random column number. Example:


SQL> SELECT * FROM (SELECT ename
  2                   FROM emp
  3                  ORDER BY dbms_random.value())
  4   WHERE rownum <= 3;
ENAME
----------
WARD
MILLER
TURNER

The ORA_HASH() function
The following example retrieves a subset of the data in the emp table by specifying 3 buckets (0 to 2) and then returning the data from bucket 1:


SELECT * FROM emp WHERE ORA_HASH(empno, 2) = 1;

How does one eliminate duplicates rows from a table?

Choose one of the following queries to identify or remove duplicate rows from a table leaving only unique records in the table:
Method 1:
Delete all rowids that is BIGGER than the SMALLEST rowid value (for a given key):


SQL> DELETE FROM table_name A
  2  WHERE ROWID > ( SELECT min(rowid)
  3                  FROM table_name B
  4                  WHERE A.key_values = B.key_values );

Method 2:
This method is usually faster. However, remember to recreate all indexes, constraints, triggers, etc. on the table when done.


SQL> create table table_name2 as select distinct * from table_name1;
SQL> drop table table_name1;
SQL> rename table_name2 to table_name1;

Method 3:


SQL> DELETE FROM my_table t1
  2  WHERE EXISTS ( SELECT 'x' FROM my_table t2
  3                 WHERE t2.key_value1 = t1.key_value1
  4                   AND t2.key_value2 = t1.key_value2
  4                   AND t2.rowid      > t1.rowid );

Note: One can eliminate N^2 unnecessary operations by creating an index on the joined fields in the inner loop (no need to loop through the entire table on each pass by a record). This will speed-up the deletion process.
Note 2: If you are comparing NULL columns, use the NVL function. Remember that NULL is not equal to NULL. This should not be a problem as all key columns should be NOT NULL by definition.
Method 4:
This method collects the first row (order by rowid) for each key values and delete the rows that are not in this set:


SQL> DELETE FROM my_table t1
  1  WHERE rowid NOT IN ( SELECT min(rowid)
  2                       FROM my_table t2
  3                       GROUP BY key_value1, key_value2 );

Note: IF key_value1 is null or key_value2 is null, this still works correctly.


How does one get the time difference between two date columns?

Oracle allows two date values to be subtracted from each other returning a numeric value indicating the number of days between the two dates (may be a fraction). This example will show how to relate it back to a time value.
Let's investigate some solutions. Test data:


SQL> CREATE TABLE dates (date1 DATE, date2 DATE);
Table created.
SQL>
SQL> INSERT INTO dates VALUES (SYSDATE, SYSDATE-1);
1 row created.
SQL> INSERT INTO dates VALUES (SYSDATE, SYSDATE-1/24);
1 row created.
SQL> INSERT INTO dates VALUES (SYSDATE, SYSDATE-1/60/24);
1 row created.
SQL> SELECT (date1 - date2) FROM dates;
DATE1-DATE2
-----------
          1
 .041666667
 .000694444

Solution 1


SQL> SELECT floor(((date1-date2)*24*60*60)/3600)
  2         || ' HOURS ' ||
  3         floor((((date1-date2)*24*60*60) -
  4         floor(((date1-date2)*24*60*60)/3600)*3600)/60)
  5         || ' MINUTES ' ||
  6         round((((date1-date2)*24*60*60) -
  7         floor(((date1-date2)*24*60*60)/3600)*3600 -
  8         (floor((((date1-date2)*24*60*60) -
  9         floor(((date1-date2)*24*60*60)/3600)*3600)/60)*60) ))
 10         || ' SECS ' time_difference
 11    FROM dates;
TIME_DIFFERENCE
--------------------------------------------------------------------------------
24 HOURS 0 MINUTES 0 SECS
1 HOURS 0 MINUTES 0 SECS
0 HOURS 1 MINUTES 0 SECS

Solution 2
If you don't want to go through the floor and ceiling maths, try this method:


SQL> SELECT to_number( to_char(to_date('1','J') +
  2         (date1 - date2), 'J') - 1)  days,
  3         to_char(to_date('00:00:00','HH24:MI:SS') +
  4         (date1 - date2), 'HH24:MI:SS') time
  5   FROM dates;
      DAYS TIME
---------- --------
         1 00:00:00
         0 01:00:00
         0 00:01:00

Solution 3
Here is a simpler method:


SQL> SELECT trunc(date1-date2) days,
  2     to_char(trunc(sysdate) + (date1 - date2),
  3             'HH24 "Hours" MI "Minutes" SS "Seconds"') time
  4   FROM dates;
      DAYS TIME
---------- ------------------------------
         1 00 Hours 00 Minutes 00 Seconds
         0 01 Hours 00 Minutes 00 Seconds
         0 00 Hours 01 Minutes 00 Seconds

How does one add a day/hour/minute/second to a date value?

The SYSDATE pseudo-column shows the current system date and time. Adding 1 to SYSDATE will advance the date by 1 day. Use fractions to add hours, minutes or seconds to the date. Look at these examples:


SQL> select sysdate, sysdate+1/24, sysdate +1/1440, sysdate + 1/86400 from dual;
SYSDATE              SYSDATE+1/24         SYSDATE+1/1440       SYSDATE+1/86400
-------------------- -------------------- -------------------- --------------------
03-Jul-2002 08:32:12 03-Jul-2002 09:32:12 03-Jul-2002 08:33:12 03-Jul-2002 08:32:13

The following format is frequently used with Oracle Replication:


select sysdate NOW, sysdate+30/(24*60*60) NOW_PLUS_30_SECS from dual;
NOW                  NOW_PLUS_30_SECS
-------------------- --------------------
03-JUL-2005 16:47:23 03-JUL-2005 16:47:53

Here are a couple of examples:


Description

Date Expression

Now

SYSDATE

Tomorow/ next day

SYSDATE + 1

Seven days from now

SYSDATE + 7

One hour from now

SYSDATE + 1/24

Three hours from now

SYSDATE + 3/24

A half hour from now

SYSDATE + 1/48

10 minutes from now

SYSDATE + 10/1440

30 seconds from now

SYSDATE + 30/86400

Tomorrow at 12 midnight

TRUNC(SYSDATE + 1)

Tomorrow at 8 AM

TRUNC(SYSDATE + 1) + 8/24

Next Monday at 12:00 noon

NEXT_DAY(TRUNC(SYSDATE), 'MONDAY') + 12/24

First day of the month at 12 midnight

TRUNC(LAST_DAY(SYSDATE ) + 1)

The next Monday, Wednesday or Friday at 9 a.m

TRUNC(LEAST(NEXT_DAY(sysdate, 'MONDAY'), NEXT_DAY(sysdate, 'WEDNESDAY'), NEXT_DAY(sysdate, 'FRIDAY'))) + 9/24

How does one code a matrix/crosstab/pivot report in SQL?

Newbies frequently ask how one can display "rows as columns" or "columns as rows". Look at these example crosstab queries (also sometimes called transposed, matrix or pivot queries):


SELECT  *
  FROM  (SELECT job,
                sum(decode(deptno,10,sal)) DEPT10,
                sum(decode(deptno,20,sal)) DEPT20,
                sum(decode(deptno,30,sal)) DEPT30,
                sum(decode(deptno,40,sal)) DEPT40
           FROM scott.emp
       GROUP BY job)
ORDER BY 1;
JOB           DEPT10     DEPT20     DEPT30     DEPT40
--------- ---------- ---------- ---------- ----------
ANALYST                    6000
CLERK           1300       1900        950
MANAGER         2450       2975       2850
PRESIDENT       5000
SALESMAN                              5600

Here is the same query with some fancy headers and totals:


SQL> ttitle "Crosstab Report"
SQL> break on report;
SQL> compute sum of dept10 dept20 dept30 dept40 total on report;
SQL>
SQL> SELECT     *
  2    FROM     (SELECT job,
  3                  sum(decode(deptno,10,sal)) DEPT10,
  4                  sum(decode(deptno,20,sal)) DEPT20,
  5                  sum(decode(deptno,30,sal)) DEPT30,
  6                  sum(decode(deptno,40,sal)) DEPT40,
  7                  sum(sal)                   TOTAL
  8             FROM emp
  9            GROUP BY job)
 10  ORDER BY 1;

Mon Aug 23                                                             page    1
                                Crosstab Report

JOB           DEPT10     DEPT20     DEPT30     DEPT40      TOTAL
--------- ---------- ---------- ---------- ---------- ----------
ANALYST                    6000                             6000
CLERK           1300       1900        950                  4150
MANAGER         2450       2975       2850                  8275
PRESIDENT       5000                                        5000
SALESMAN                              5600                  5600
          ---------- ---------- ---------- ---------- ----------
sum             8750      10875       9400                 29025

Here's another variation on the theme:


SQL> SELECT DECODE(MOD(v.row#,3)
  2                 ,1, 'Number: '  ||deptno
  3                 ,2, 'Name: '    ||dname
  4                 ,0, 'Location: '||loc
  5                 ) AS "DATA"
  6    FROM dept,
  7         (SELECT rownum AS row# FROM user_objects WHERE rownum < 4) v
  8   WHERE deptno = 30
  9  /
DATA
--------------------------------------- ---------
Number: 30
Name: SALES
Location: CHICAGO

From Oracle 11g, we can use pivot option


Can one retrieve only rows X to Y from a table?

SELECT * FROM (
   SELECT ename, rownum rn
            FROM emp WHERE rownum < 101
) WHERE  RN between 91 and 100 ;

Note: the 101 is just one greater than the maximum row of the required rows (means x= 90, y=100, so the inner values is y+1).


SELECT rownum, f1 FROM t1
GROUP BY rownum, f1 HAVING rownum BETWEEN 2 AND 4;

Another solution is to use the MINUS operation. For example, to display rows 5 to 7, construct a query like this:


SELECT *
FROM   tableX
WHERE  rowid in (
   SELECT rowid FROM tableX
    WHERE rownum <= 7
  MINUS
   SELECT rowid FROM tableX
   WHERE rownum < 5);

"this one was faster for me and allowed for sorting before filtering by rownum. The inner query (table A) can be a series of tables joined together with any operation before the filtering by rownum is applied."


SELECT *
  FROM (SELECT a.*, rownum RN
                       FROM (SELECT *
                  FROM t1 ORDER BY key_column) a
         WHERE rownum <=7)
 WHERE rn >=5;

Please note, there is no explicit row order in a relational database. However, this query is quite fun and may even help in the odd situation.
The generic solution to get full information of rows between x and y

SELECT * FROM emp WHERE empno in (SELECT empno FROM emp GROUP BY rownum,empno HAVING rownum BETWEEN &x AND &y);

"select particular rows from a table : select for rownum = 4, 15 and 17."


select * from (
        select rownum myrownum, emp.* from employees emp
        ) mytable
        where myrownum in (4,15,17);


"selecting row between range of rownum: select for rownum between (12, 20)."


select * from (
        select rownum myrownum, emp.* from employees emp
        ) mytable
        where myrownum between 12 and 20;


"Replace 12 and 20 with &x and &y respectively to assign range dynamically."


select * from (
        select rownum myrownum, emp.* from employees emp
        ) mytable
        where myrownum between &x and &y;


"Combined query to give complete flexibility to pick particular rows and also a given range."


select * from (
        select rownum myrownum, emp.* from employees emp
        ) mytable
        where myrownum between 12 and 17
        or myrownum in ( 3, 18, 25);

Can one retrieve only the Nth row from a table?

SELECT * FROM t1 a
WHERE  n = (SELECT COUNT(rowid)
              FROM t1 b
             WHERE a.rowid >= b.rowid);
SELECT * FROM (
   SELECT ENAME,ROWNUM RN FROM EMP WHERE ROWNUM < 101 )
 WHERE  RN = 100;

Note: In this first query we select one more than the required row number, then we select the required one. Its far better than using a MINUS operation.


SELECT f1 FROM t1
WHERE  rowid = (
        SELECT rowid FROM t1
        WHERE  rownum <= 10
        MINUS
        SELECT rowid FROM t1
        WHERE  rownum < 10);
SELECT rownum,empno FROM scott.emp a
 GROUP BY rownum,empno HAVING rownum = 4;

Alternatively...


SELECT * FROM emp WHERE rownum=1 AND rowid NOT IN
  (SELECT rowid FROM emp WHERE rownum < 10);

Please note, there is no explicit row order in a relational database. However, this query is quite fun and may even help in the odd situation.


How can one dump/ examine the exact content of a database column?

Table data can be extracted from the database as octal, decimal or hex values:


SELECT DUMP(col1, 10)
FROM tab1
WHERE cond1 = val1;
DUMP(COL1)
-------------------------------------
Typ=96 Len=4: 65,66,67,32

For this example, type=96 is indicating a CHAR column. The last byte in the column is 32, which is the ASCII code for a space. This tells us that this column is blank-padded.


How does one add a column to the middle of a table?

Oracle only allows columns to be added to the end of an existing table. Example:


SQL> CREATE TABLE tab1 ( col1 NUMBER );
Table created.

SQL> ALTER TABLE tab1 ADD (col2 DATE);
Table altered.

SQL> DESC tab1
Name                                      Null?    Type
----------------------------------------- -------- ----------------------------
COL1                                               NUMBER
COL2                                               DATE

Nevertheless, some databases also allow columns to be added to an existing table after a particular column (i.e. in the middle of the table). For example, in MySQL the following syntax is valid:


ALTER TABLE tablename ADD columnname AFTER columnname;

Oracle does not support this syntax. However, it doesn't mean that it cannot be done.
Workarounds:
1. Create a new table and copy the data across.


SQL> RENAME tab1 TO tab1_old;
Table renamed.

SQL> CREATE TABLE tab1 AS SELECT 0 AS col1, col1 AS col2 FROM tab1_old;
Table created.

2. Rename the table and create a view upon it with its former name and with the columns in the order you want.
3. Use the DBMS_REDEFINITION package to change the structure on-line while users are working.


How does one code a hierarchical tree-structured query?

The SCOTT/TIGER database schema contains a table EMP with a self-referencing relation (EMPNO and MGR columns). This table is perfect for testing and demonstrating tree-structured queries as the MGR column contains the employee number of the "current" employee's boss.
The LEVEL pseudo-column is an indication of how deep in the tree one is. Oracle can handle queries with a depth of up to 255 levels. Look at this example:


SQL> SELECT     level, empno, ename, mgr
  2    FROM     emp
  3  CONNECT BY PRIOR empno = mgr
  4    START WITH mgr IS NULL
  5  /
     LEVEL      EMPNO ENAME             MGR
---------- ---------- ---------- ----------
         1       7839 KING
         2       7566 JONES            7839
         3       7788 SCOTT            7566
...

One can produce an indented report by using the level number to substring or lpad() a series of spaces, and concatenate that to the string. Look at this example:


SQL> SELECT     LPAD(' ', LEVEL * 2) || ename
  2    FROM     emp
  3  CONNECT BY PRIOR empno = mgr
  4    START WITH mgr IS NULL;
LPAD(,LEVEL*2)||ENAME
------------------------------------------------------
  KING
    JONES
      SCOTT
...

Use the "start with" clause to specify the start of the tree. More than one record can match the starting condition. One disadvantage of having a "connect by prior" clause is that you cannot perform a join to other tables. The "connect by prior" clause is rarely implemented in the other database offerings. Trying to do this programmatically is difficult as one has to do the top level query first, then, for each of the records open a cursor to look for child nodes.
One way of working around this is to use PL/SQL, open the driving cursor with the "connect by prior" statement, and the select matching records from other tables on a row-by-row basis, inserting the results into a temporary table for later retrieval.
NOTE: Tree-structured queries are definitely non-relational (enough to kill Codd and make him roll in his grave). Also, this feature is not often found in other database offerings.


How does one count/sum data values in a column?

Count/sum FIX values:
Use this simple query to count the number of data values in a column:


select my_table_column, count(*)
from   my_table
group  by my_table_column;

A more sophisticated example...


select dept, count(decode(sex,'M',1)) MALE,
             count(decode(sex,'F',1)) FEMALE,
             count(decode(sex,'M',null,'F',null,1)) OTHER,
             count(*) TOTAL
  from my_emp_table
 group by dept;

Count/sum RANGES of data values in a column:
A value x will be between values y and z if GREATEST(x, y) = LEAST(x, z). Look at this example:


select f2,
       sum(decode(greatest(f1,59), least(f1,100), 1, 0)) "Range 60-100",
       sum(decode(greatest(f1,30), least(f1, 59), 1, 0)) "Range 30-59",
       sum(decode(greatest(f1, 0), least(f1, 29), 1, 0)) "Range 00-29"
from   my_table
group  by f2;

For equal size ranges it might be easier to calculate it with DECODE(TRUNC(value/range), 0, rate_0, 1, rate_1, ...). Eg.


select ename "Name", sal "Salary",
       decode( trunc(f2/1000, 0), 0, 0.0,
                                  1, 0.1,
                                  2, 0.2,
                                  3, 0.31) "Tax rate"
from   my_table;

How does one drop/ rename a column in a table?

Drop a column
From Oracle 8i one can DROP a column from a table. Look at this sample script, demonstrating the ALTER TABLE table_name DROP COLUMN column_name; command.
Workarounds for older releases:


SQL> update t1 set column_to_drop = NULL;
SQL> rename t1 to t1_base;
SQL> create view t1 as select >specific columns> from t1_base;
SQL> create table t2 as select >specific columns> from t1;
SQL> drop table t1;
SQL> rename t2 to t1;

Rename a column
From Oracle 9i one can RENAME a column from a table. Look at this example:


ALTER TABLE tablename RENAME COLUMN oldcolumn TO newcolumn;

Workarounds for older releases:
Use a view with correct column names:


rename t1 to t1_base;
create view t1 >column list with new name> as select * from t1_base;

Recreate the table with correct column names:


create table t2 >column list with new name> as select * from t1;
drop table t1;
rename t2 to t1;

Add a column with a new name and drop an old column:


alter table t1 add ( newcolame datatype ); 
update t1 set newcolname=oldcolname;
alter table t1 drop column oldcolname;

How does one implement IF-THEN-ELSE logic in a SELECT statement?

One can use the CASE expression or functions like DECODE, NVL, NVL2, NULLIF, COALESCE, etc.
Here is the syntax for the CASE-statement:


CASE exp WHEN comparison_exp1 THEN return_exp1
        [WHEN comparison_exp2 THEN return_exp2
         WHEN comparison_exp3 THEN return_exp3
          ELSE else_exp
        ]
END

And for DECODE:


DECODE( col | exprn, srch1, rslt1 [, srch2, rslt2,...,] [,default] )

How does one prevent Oracle from using an Index?

In certain cases, one may want to disable the use of a specific, or all indexes for a given query. Here are some examples:
Adding an expression to the indexed column


SQL>select count(*) from t where empno+0=1000;
  COUNT(*)
----------
         1

Execution Plan
--------------------------------------------- ----- --------
   0      SELECT STATEMENT Optimizer=CHOOSE (Cost=2 Card=1 Bytes=3)
   1    0   SORT (AGGREGATE)
   2    1     TABLE ACCESS (FULL) OF 'T' (Cost=2 Card=1 Bytes=3)

Specifying the FULL hint to force full table scan


SQL>select /*+ FULL(t) */ * from t where empno=1000;
     EMPNO ENAME      JOB              MGR HIREDATE         SAL       COMM     DEPTNO GRADE
---------- ---------- --------- ---------- --------- ---------- ---------- ---------- ----------
      1000 Victor     DBA             7839 20-MAY-03      11000          0         10 JUNIOR

Execution Plan
--------------------------------------------- ----- --------
   0      SELECT STATEMENT Optimizer=CHOOSE (Cost=2 Card=1 Bytes=41)
   1    0   TABLE ACCESS (FULL) OF 'T' (Cost=2 Card=1 Bytes=41)

Specifying NO_INDEX hint


SQL>select /*+ NO_INDEX(T) */ count(*) from t where empno=1000;
  COUNT(*)
----------
         1

Execution Plan
--------------------------------------------- ----- --------
   0      SELECT STATEMENT Optimizer=CHOOSE (Cost=2 Card=1 Bytes=3)
   1    0   SORT (AGGREGATE)
   2    1     TABLE ACCESS (FULL) OF 'T' (Cost=2 Card=1 Bytes=3)

Using a function over the indexed column


SQL>select count(*) from t where to_number(empno)=1000;
  COUNT(*)
----------
         1

Execution Plan
--------------------------------------------- ----- --------
   0      SELECT STATEMENT Optimizer=CHOOSE (Cost=2 Card=1 Bytes=3)
   1    0   SORT (AGGREGATE)
   2    1     TABLE ACCESS (FULL) OF 'T' (Cost=2 Card=1 Bytes=3)

How does one select EVERY Nth row from a table?

One can easily select all even, odd, or Nth rows from a table using SQL queries like this:
Method 1: Using a subquery


SELECT *
FROM   emp
WHERE  (ROWID,0) IN (SELECT ROWID, MOD(ROWNUM,4)
                     FROM   emp);

Method 2: Use dynamic views (available from Oracle7.2):


SELECT *
FROM   ( SELECT rownum rn, empno, ename
         FROM emp
       ) temp
WHERE  MOD(temp.ROWNUM,4) = 0;

Method 3: Using GROUP BY and HAVING


SELECT rownum, f1 FROM t1
GROUP BY rownum, f1 HAVING MOD(rownum,n) = 0 OR rownum = 2-n;

Please note, there is no explicit row order in a relational database. However, these queries are quite fun and may even help in the odd situation.


How does one select the LAST N rows from a table?

From Oracle 9i onwards, the RANK() and DENSE_RANK() functions can be used to determine theLAST N or BOTTOM N rows. Examples:
Get the bottom 10 employees based on their salary


SELECT ename, sal
  FROM ( SELECT ename, sal, RANK() OVER (ORDER BY sal ASC) sal_rank
           FROM emp )
 WHERE sal_rank <= 10;

Select the employees getting the lowest 10 salaries


SELECT ename, sal
  FROM ( SELECT ename, sal, DENSE_RANK() OVER (ORDER BY sal) sal_dense_rank
           FROM emp )
 WHERE sal_dense_rank <= 10;

For Oracle 8i and above, one can get the bottom N rows using an inner-query with an ORDER BY clause:


SELECT *
  FROM (SELECT * FROM my_table ORDER BY col_name_1)
 WHERE ROWNUM < 10;

Use this workaround for older (8.0 and prior) releases:


SELECT *
  FROM my_table a
 WHERE 10 >= (SELECT COUNT(DISTINCT maxcol)
                FROM my_table b
               WHERE b.maxcol <= a.maxcol)
 ORDER BY maxcol;

How does one select the TOP N rows from a table?

From Oracle 9i onwards, the RANK() and DENSE_RANK() functions can be used to determine the TOP N rows. Examples:
Get the top 10 employees based on their salary


SELECT ename, sal
  FROM ( SELECT ename, sal, RANK() OVER (ORDER BY sal DESC) sal_rank
           FROM emp )
 WHERE sal_rank <= 10;

Select the employees making the top 10 salaries


SELECT ename, sal
  FROM ( SELECT ename, sal, DENSE_RANK() OVER (ORDER BY sal DESC) sal_dense_rank
           FROM emp )
 WHERE sal_dense_rank <= 10;

For Oracle 8i and above, one can get the Top N rows using an inner-query with an ORDER BY clause:


SELECT *
  FROM (SELECT * FROM my_table ORDER BY col_name_1 DESC)
 WHERE ROWNUM < 10;

Use this workaround for older (8.0 and prior) releases:


SELECT *
  FROM my_table a
 WHERE 10 >= (SELECT COUNT(DISTINCT maxcol)
                FROM my_table b
               WHERE b.maxcol >= a.maxcol)
 ORDER BY maxcol DESC;

How to generate a text graphs (histograms) using SQL?

SELECT d.dname AS "Department",
             LPAD('+', COUNT(*), '+') as "Graph"
  FROM emp e, dept d
 WHERE e.deptno = d.deptno
 GROUP BY d.dname;

Sample output:


Department     Graph
-------------- --------------------------------------------------
ACCOUNTING     +++
RESEARCH       +++++
SALES          ++++++

In the above example, the value returned by COUNT(*) is used to control the number of "*" characters to return for each department. We simply pass COUNT(*) as an argument to the string function LPAD (or RPAD) to return the desired number of *'s.


Map/ concatenate several rows to a column

This FAQ will demonstrate how row values can be concatenated into a single column value (similar to MySQL's [i]GROUP_CONCAT[/i] function).
Start by creating this function:


SQL> CREATE OR REPLACE FUNCTION rowconcat(q in VARCHAR2) RETURN VARCHAR2 IS
  2    ret  VARCHAR2(4000);
  3    hold VARCHAR2(4000);
  4    cur  sys_refcursor;
  5  BEGIN
  6    OPEN cur FOR q;
  7    LOOP
  8      FETCH cur INTO hold;
  9      EXIT WHEN cur%NOTFOUND;
 10      IF ret IS NULL THEN
 11        ret := hold;
 12      ELSE
 13        ret := ret || ',' || hold;
 14      END IF;
 15    END LOOP;
 16    RETURN ret;
 17  END;
 18  /
Function created.

This function returns a string result with the concatenated non-NULL values from a SQL statement. It returns NULL if there are no non-NULL values.
Here is an example of how to map several rows to a single concatenated column:


SQL> SELECT rowconcat('SELECT dname FROM dept') AS departments
  2    FROM dual;
DEPARTMENTS
--------------------------------------------------------------------------------
ACCOUNTING,RESEARCH,SALES,OPERATIONS

This example is more interresting, it concatenates a column across several rows based on an aggregation:


SQL> col employees format a50
SQL> SELECT deptno,
  2         rowconcat('SELECT ename FROM emp a WHERE deptno='||deptno) AS Employees
  3    FROM emp
  4   GROUP BY deptno
  5  /
    DEPTNO EMPLOYEES
---------- --------------------------------------------------
        30 ALLEN,WARD,MARTIN,BLAKE,TURNER,JAMES
        20 SMITH,JONES,SCOTT,ADAMS,FORD
        10 CLARK,KING,MILLER

What is the difference between VARCHAR, VARCHAR2 and CHAR data types?

Both CHAR and VARCHAR2 types are used to store character string values, however, they behave very differently. The VARCHAR type should not be used:
CHAR
CHAR should be used for storing fixed length character strings. String values will be space/blank padded before stored on disk. If this type is used to store variable length strings, it will waste a lot of disk space.


SQL> CREATE TABLE char_test (col1 CHAR(10));
Table created.

SQL> INSERT INTO char_test VALUES ('qwerty');
1 row created.

SQL> SELECT col1, length(col1), dump(col1) "ASCII Dump" FROM char_test;
COL1       LENGTH(COL1) ASCII Dump
---------- ------------ ------------------------------------------------------------
qwerty               10 Typ=96 Len=10: 113,119,101,114,116,121,32,32,32,32

Note: ASCII character 32 is a blank space.
VARCHAR
Currently VARCHAR behaves exactly the same as VARCHAR2. However, this type should not be used as it is reserved for future usage.


SQL> CREATE TABLE varchar_test (col1 VARCHAR2(10));
Table created.

SQL> INSERT INTO varchar_test VALUES ('qwerty');
1 row created.

SQL> SELECT col1, length(col1), dump(col1) "ASCII Dump" FROM varchar_test;
COL1       LENGTH(COL1) ASCII Dump
---------- ------------ ------------------------------------------------------------
qwerty                6 Typ=1 Len=6: 113,119,101,114,116,121

VARCHAR2
VARCHAR2 is used to store variable length character strings. The string value's length will be stored on disk with the value itself.


SQL> CREATE TABLE varchar2_test (col1 VARCHAR2(10));
Table created.

SQL> INSERT INTO varchar2_test VALUES ('qwerty');
1 row created.

SQL> SELECT col1, length(col1), dump(col1) "ASCII Dump" FROM varchar2_test;
COL1       LENGTH(COL1) ASCII Dump
---------- ------------ ------------------------------------------------------------
qwerty                6 Typ=1 Len=6: 113,119,101,114,116,121

1. Which is the subset of SQL commands used to manipulate Oracle Database structures, including tables?
Data Definition Language (DDL)

2. What operator performs pattern matching?
LIKE operator

3. What operator tests column for the absence of data?
IS NULL operator

4. Which command executes the contents of a specified file?
START <filename> or @<filename>

5. What is the parameter substitution symbol used with INSERT INTO command?
&

6. Which command displays the SQL command in the SQL buffer, and then executes it?
RUN

7. What are the wildcards used for pattern matching?
_ for single character substitution and % for multi-character substitution

8. State true or false. EXISTS, SOME, ANY are operators in SQL.
True

9. State true or false. !=, <>, ^= all denote the same operation.
True

10. What are the privileges that can be granted on a table by a user to others?
Insert, update, delete, select, references, index, execute, alter, all

11. What command is used to get back the privileges offered by the GRANT command?
REVOKE

12. Which system tables contain information on privileges granted and privileges obtained?
USER_TAB_PRIVS_MADE, USER_TAB_PRIVS_RECD

13. Which system table contains information on constraints on all the tables created?
USER_CONSTRAINTS

14. TRUNCATE TABLE EMP;
DELETE FROM EMP;
Will the outputs of the above two commands differ?
Both will result in deleting all the rows in the table EMP.

15. What is the difference between TRUNCATE and DELETE commands?
TRUNCATE is a DDL command whereas DELETE is a DML command. Hence DELETE operation can be rolled back, but TRUNCATE operation cannot be rolled back. WHERE clause can be used with DELETE and not with TRUNCATE.

16. What command is used to create a table by copying the structure of another table?
Answer :
CREATE TABLE .. AS SELECT command
Explanation :
To copy only the structure, the WHERE clause of the SELECT command should contain a FALSE statement as in the following.
CREATE TABLE NEWTABLE AS SELECT * FROM EXISTINGTABLE WHERE 1=2;
If the WHERE condition is true, then all the rows or rows satisfying the condition will be copied to the new table.

17. What will be the output of the following query?
SELECT REPLACE(TRANSLATE(LTRIM(RTRIM('!! ATHEN !!','!'), '!'), 'AN', '**'),'*','TROUBLE') FROM DUAL;
TROUBLETHETROUBLE

18. What will be the output of the following query?
SELECT DECODE(TRANSLATE('A','1234567890','1111111111'), '1','YES', 'NO' );
Answer :
NO
Explanation :
The query checks whether a given string is a numerical digit.

19. What does the following query do?
SELECT SAL + NVL(COMM,0) FROM EMP;
This displays the total salary of all employees. The null values in the commission column will be replaced by 0 and added to salary.

20. Which date function is used to find the difference between two dates?
MONTHS_BETWEEN

21. Why does the following command give a compilation error?
DROP TABLE &TABLE_NAME;
Variable names should start with an alphabet. Here the table name starts with an '&' symbol.

22. What is the advantage of specifying WITH GRANT OPTION in the GRANT command?
The privilege receiver can further grant the privileges he/she has obtained from the owner to any other user.

23. What is the use of the DROP option in the ALTER TABLE command?
It is used to drop constraints specified on the table.

24. What is the value of ‘comm’ and ‘sal’ after executing the following query if the initial value of ‘sal’ is 10000?
UPDATE EMP SET SAL = SAL + 1000, COMM = SAL*0.1;
sal = 11000, comm = 1000

25. What is the use of DESC in SQL?
Answer :
DESC has two purposes. It is used to describe a schema as well as to retrieve rows from table in descending order.
Explanation :
The query SELECT * FROM EMP ORDER BY ENAME DESC will display the output sorted on ENAME in descending order.

26. What is the use of CASCADE CONSTRAINTS?
When this clause is used with the DROP command, a parent table can be dropped even when a child table exists.

27. Which function is used to find the largest integer less than or equal to a specific value?
FLOOR

28. What is the output of the following query?
SELECT TRUNC(1234.5678,-2) FROM DUAL;
1200


SQL – QUERIES

I. SCHEMAS

Table 1 : STUDIES

PNAME (VARCHAR), SPLACE (VARCHAR), COURSE (VARCHAR), CCOST (NUMBER)

Table 2 : SOFTWARE

PNAME (VARCHAR), TITLE (VARCHAR), DEVIN (VARCHAR), SCOST (NUMBER), DCOST (NUMBER), SOLD (NUMBER)

Table 3 : PROGRAMMER

PNAME (VARCHAR), DOB (DATE), DOJ (DATE), SEX (CHAR), PROF1 (VARCHAR), PROF2 (VARCHAR), SAL (NUMBER)

LEGEND :

PNAME – Programmer Name, SPLACE – Study Place, CCOST – Course Cost, DEVIN – Developed in, SCOST – Software Cost, DCOST – Development Cost, PROF1 – Proficiency 1

QUERIES :

1. Find out the selling cost average for packages developed in Oracle.
2. Display the names, ages and experience of all programmers.
3. Display the names of those who have done the PGDCA course.
4. What is the highest number of copies sold by a package?
5. Display the names and date of birth of all programmers born in April.
6. Display the lowest course fee.
7. How many programmers have done the DCA course.
8. How much revenue has been earned through the sale of packages developed in C.
9. Display the details of software developed by Rakesh.
10. How many programmers studied at Pentafour.
11. Display the details of packages whose sales crossed the 5000 mark.
12. Find out the number of copies which should be sold in order to recover the development cost of each package.
13. Display the details of packages for which the development cost has been recovered.
14. What is the price of costliest software developed in VB?
15. How many packages were developed in Oracle ?
16. How many programmers studied at PRAGATHI?
17. How many programmers paid 10000 to 15000 for the course?
18. What is the average course fee?
19. Display the details of programmers knowing C.
20. How many programmers know either C or Pascal?
21. How many programmers don’t know C and C++?
22. How old is the oldest male programmer?
23. What is the average age of female programmers?
24. Calculate the experience in years for each programmer and display along with their names in descending order.
25. Who are the programmers who celebrate their birthdays during the current month?
26. How many female programmers are there?
27. What are the languages known by the male programmers?
28. What is the average salary?
29. How many people draw 5000 to 7500?
30. Display the details of those who don’t know C, C++ or Pascal.
31. Display the costliest package developed by each programmer.
32. Produce the following output for all the male programmers
Programmer
Mr. Arvind – has 15 years of experience

KEYS:

1. SELECT AVG(SCOST) FROM SOFTWARE WHERE DEVIN = 'ORACLE';
2. SELECT PNAME,TRUNC(MONTHS_BETWEEN(SYSDATE,DOB)/12) "AGE", TRUNC(MONTHS_BETWEEN(SYSDATE,DOJ)/12) "EXPERIENCE" FROM PROGRAMMER;
3. SELECT PNAME FROM STUDIES WHERE COURSE = 'PGDCA';
4. SELECT MAX(SOLD) FROM SOFTWARE;
5. SELECT PNAME, DOB FROM PROGRAMMER WHERE DOB LIKE '%APR%';
6. SELECT MIN(CCOST) FROM STUDIES;
7. SELECT COUNT(*) FROM STUDIES WHERE COURSE = 'DCA';
8. SELECT SUM(SCOST*SOLD-DCOST) FROM SOFTWARE GROUP BY DEVIN HAVING DEVIN = 'C';
9. SELECT * FROM SOFTWARE WHERE PNAME = 'RAKESH';
10. SELECT * FROM STUDIES WHERE SPLACE = 'PENTAFOUR';
11. SELECT * FROM SOFTWARE WHERE SCOST*SOLD-DCOST > 5000;
12. SELECT CEIL(DCOST/SCOST) FROM SOFTWARE;
13. SELECT * FROM SOFTWARE WHERE SCOST*SOLD >= DCOST;
14. SELECT MAX(SCOST) FROM SOFTWARE GROUP BY DEVIN HAVING DEVIN = 'VB';
15. SELECT COUNT(*) FROM SOFTWARE WHERE DEVIN = 'ORACLE';
16. SELECT COUNT(*) FROM STUDIES WHERE SPLACE = 'PRAGATHI';
17. SELECT COUNT(*) FROM STUDIES WHERE CCOST BETWEEN 10000 AND 15000;
18. SELECT AVG(CCOST) FROM STUDIES;
19. SELECT * FROM PROGRAMMER WHERE PROF1 = 'C' OR PROF2 = 'C';
20. SELECT * FROM PROGRAMMER WHERE PROF1 IN ('C','PASCAL') OR PROF2 IN ('C','PASCAL');
21. SELECT * FROM PROGRAMMER WHERE PROF1 NOT IN ('C','C++') AND PROF2 NOT IN ('C','C++');
22. SELECT TRUNC(MAX(MONTHS_BETWEEN(SYSDATE,DOB)/12)) FROM PROGRAMMER WHERE SEX = 'M';
23. SELECT TRUNC(AVG(MONTHS_BETWEEN(SYSDATE,DOB)/12)) FROM PROGRAMMER WHERE SEX = 'F';
24. SELECT PNAME, TRUNC(MONTHS_BETWEEN(SYSDATE,DOJ)/12) FROM PROGRAMMER ORDER BY PNAME DESC;
25. SELECT PNAME FROM PROGRAMMER WHERE TO_CHAR(DOB,'MON') = TO_CHAR(SYSDATE,'MON');
26. SELECT COUNT(*) FROM PROGRAMMER WHERE SEX = 'F';
27. SELECT DISTINCT(PROF1) FROM PROGRAMMER WHERE SEX = 'M';
28. SELECT AVG(SAL) FROM PROGRAMMER;
29. SELECT COUNT(*) FROM PROGRAMMER WHERE SAL BETWEEN 5000 AND 7500;
30. SELECT * FROM PROGRAMMER WHERE PROF1 NOT IN ('C','C++','PASCAL') AND PROF2 NOT IN ('C','C++','PASCAL');
31. SELECT PNAME,TITLE,SCOST FROM SOFTWARE WHERE SCOST IN (SELECT MAX(SCOST) FROM SOFTWARE GROUP BY PNAME);
32.SELECT 'Mr.' || PNAME || ' - has ' || TRUNC(MONTHS_BETWEEN(SYSDATE,DOJ)/12) || ' years of experience' “Programmer” FROM PROGRAMMER WHERE SEX = 'M' UNION SELECT 'Ms.' || PNAME || ' - has ' || TRUNC (MONTHS_BETWEEN (SYSDATE,DOJ)/12) || ' years of experience' “Programmer” FROM PROGRAMMER WHERE SEX = 'F';

 

II . SCHEMA :

Table 1 : DEPT

DEPTNO (NOT NULL , NUMBER(2)), DNAME (VARCHAR2(14)),
LOC (VARCHAR2(13)

Table 2 : EMP

EMPNO (NOT NULL , NUMBER(4)), ENAME (VARCHAR2(10)),
JOB (VARCHAR2(9)), MGR (NUMBER(4)), HIREDATE (DATE),
SAL (NUMBER(7,2)), COMM (NUMBER(7,2)), DEPTNO (NUMBER(2))

MGR is the empno of the employee whom the employee reports to. DEPTNO is a foreign key.
QUERIES

1. List all the employees who have at least one person reporting to them.
2. List the employee details if and only if more than 10 employees are present in department no 10.
3. List the name of the employees with their immediate higher authority.
4. List all the employees who do not manage any one.
5. List the employee details whose salary is greater than the lowest salary of an employee belonging to deptno 20.
6. List the details of the employee earning more than the highest paid manager.
7. List the highest salary paid for each job.
8. Find the most recently hired employee in each department.
9. In which year did most people join the company? Display the year and the number of employees.
10. Which department has the highest annual remuneration bill?
11. Write a query to display a ‘*’ against the row of the most recently hired employee.
12. Write a correlated sub-query to list out the employees who earn more than the average salary of their department.
13. Find the nth maximum salary.
14. Select the duplicate records (Records, which are inserted, that already exist) in the EMP table.
15. Write a query to list the length of service of the employees (of the form n years and m months).

KEYS:

1. SELECT DISTINCT(A.ENAME) FROM EMP A, EMP B WHERE A.EMPNO = B.MGR; or SELECT ENAME FROM EMP WHERE EMPNO IN (SELECT MGR FROM EMP);
2. SELECT * FROM EMP WHERE DEPTNO IN (SELECT DEPTNO FROM EMP GROUP BY DEPTNO HAVING COUNT(EMPNO)>10 AND DEPTNO=10);
3. SELECT A.ENAME "EMPLOYEE", B.ENAME "REPORTS TO" FROM EMP A, EMP B WHERE A.MGR=B.EMPNO;
4. SELECT * FROM EMP WHERE EMPNO IN ( SELECT EMPNO FROM EMP MINUS SELECT MGR FROM EMP);
5. SELECT * FROM EMP WHERE SAL > ( SELECT MIN(SAL) FROM EMP GROUP BY DEPTNO HAVING DEPTNO=20);
6. SELECT * FROM EMP WHERE SAL > ( SELECT MAX(SAL) FROM EMP GROUP BY JOB HAVING JOB = 'MANAGER' );
7. SELECT JOB, MAX(SAL) FROM EMP GROUP BY JOB;
8. SELECT * FROM EMP WHERE (DEPTNO, HIREDATE) IN (SELECT DEPTNO, MAX(HIREDATE) FROM EMP GROUP BY DEPTNO);
9. SELECT TO_CHAR(HIREDATE,'YYYY') "YEAR", COUNT(EMPNO) "NO. OF EMPLOYEES" FROM EMP GROUP BY TO_CHAR(HIREDATE,'YYYY') HAVING COUNT(EMPNO) = (SELECT MAX(COUNT(EMPNO)) FROM EMP GROUP BY TO_CHAR(HIREDATE,'YYYY'));
10. SELECT DEPTNO, LPAD(SUM(12*(SAL+NVL(COMM,0))),15) "COMPENSATION" FROM EMP GROUP BY DEPTNO HAVING SUM( 12*(SAL+NVL(COMM,0))) = (SELECT MAX(SUM(12*(SAL+NVL(COMM,0)))) FROM EMP GROUP BY DEPTNO);
11. SELECT ENAME, HIREDATE, LPAD('*',8) "RECENTLY HIRED" FROM EMP WHERE HIREDATE = (SELECT MAX(HIREDATE) FROM EMP) UNION SELECT ENAME NAME, HIREDATE, LPAD(' ',15) "RECENTLY HIRED" FROM EMP WHERE HIREDATE != (SELECT MAX(HIREDATE) FROM EMP);
12. SELECT ENAME,SAL FROM EMP E WHERE SAL > (SELECT AVG(SAL) FROM EMP F WHERE E.DEPTNO = F.DEPTNO);
13. SELECT ENAME, SAL FROM EMP A WHERE &N = (SELECT COUNT (DISTINCT(SAL)) FROM EMP B WHERE A.SAL<=B.SAL);
14. SELECT * FROM EMP A WHERE A.EMPNO IN (SELECT EMPNO FROM EMP GROUP BY EMPNO HAVING COUNT(EMPNO)>1) AND A.ROWID!=MIN (ROWID));
15. SELECT ENAME "EMPLOYEE",TO_CHAR(TRUNC(MONTHS_BETWEEN(SYSDATE,HIREDATE)/12))||' YEARS '|| TO_CHAR(TRUNC(MOD(MONTHS_BETWEEN (SYSDATE, HIREDATE),12)))||' MONTHS ' "LENGTH OF SERVICE" FROM EMP;

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