Database: SQL INTERVIEW QUES

Showing posts with label SQL INTERVIEW QUES. Show all posts

Saturday, 28 April 2012

Delete duplicate values from a table

DELETE FROM my_table

 WHERE ROWID NOT IN (SELECT MIN(ROWID)

                       FROM my_table

                      GROUP BY delete_col_name);

-- Example :

--

-- Given a table called emp with the following columns:

--   id   number

--   name varchar2(20)

--   sal  number

--

-- To delete the duplicate values:

--

-- DELETE FROM emp

--  WHERE ROWID NOT IN (SELECT MIN(ROWID) FROM emp GROUP BY id);

--

-- COMMIT;

Convert LONG data types to LOBs

create table old_long_table(c1 number, c2 long);

insert into  old_long_table values (1, 'LONG data to convert to CLOB');

create table new_lob_table(c1 number, c2 clob);

-- Use TO_LOB function to convert LONG to LOB...

insert into  new_lob_table

       select c1, to_lob(c2) from old_long_table;

-- Note: the same procdure can be used to convert LONG RAW datatypes to BLOBs.

Demonstrate Oracle temporary tables

drop table x

-------------------------

create global temporary table x (a date)

        on commit delete rows     -- Delete rows after commit

        -- on commit preserve rows   -- Delete rows after exit session

-------------------------

select table_name, temporary, duration

from   user_tables

where  table_name = 'X'

--------------------------

insert into x values (sysdate);

select * from x;

commit;

-- Inserted rows are missing after commit

select * from x;

Demonstrate VARRAY database types

CREATE OR REPLACE TYPE vcarray AS VARRAY(10) OF VARCHAR2(128);

-----------------

CREATE TABLE varray_table (id number, col1 vcarray);

INSERT INTO varray_table VALUES (1, vcarray('A'));

INSERT INTO varray_table VALUES (2, vcarray('B', 'C'));

INSERT INTO varray_table VALUES (3, vcarray('D', 'E', 'F'));

SELECT * FROM varray_table;

SELECT * FROM USER_VARRAYS;

-- SELECT * FROM USER_SEGMENTS;

-- Unnesting the collection:

select t1.id, t2.COLUMN_VALUE

from   varray_table t1, TABLE(t1.col1) t2

 -----------------------------

-- Use PL/SQL to access the varray...

set serveroutput on

declare

  v_vcarray vcarray;

begin

  for c1 in (select * from varray_table) loop

      dbms_output.put_line('Row fetched...');

      FOR i IN c1.col1.FIRST..c1.col1.LAST LOOP

          dbms_output.put_line('...property fetched: '|| c1.col1(i));

      END LOOP;

  end loop;

end;

------------------------

 -- Clean-up...

DROP TABLE varray_table;

Demonstrate Oracle database types and object tables

drop type employee_typ;

create type employee_typ as object (

        empno NUMBER,

        emp_name varchar2(30),

        hiredate date,

        member function days_at_company return NUMBER,

        pragma restrict_references(days_at_company, WNDS)

------------

create type body employee_tye is

begin

        member function days_at_company return number is

        begin

                return (SYSDATE-hiredate);

        end;

end;

Count the number of rows for ALL tables in current schema

set termout off echo off feed off trimspool on head off pages 0

spool countall.tmp

select 'SELECT count(*), '''||table_name||''' from '||table_name||';'

from   user_tables

spool off

set termout on

@@countall.tmp

set head on feed on

SQL*Plus Help script,Test for Leap Years, Spell out numbers to words,Demonstrate simple encoding and decoding of messages

SQL*Plus Help script

select info from system.help where upper(topic)=upper('&1')

Test for Leap Years

select year,

decode( mod(year, 4), 0,

decode( mod(year, 400), 0, 'Leap Year',

decode( mod(year, 100), 0, 'Not a Leap Year', 'Leap Year')

), 'Not a Leap Year'

) as leap_year_indicator

from my_table

Spell out numbers to words

select decode( sign( &num ), -1, 'Negative ', 0, 'Zero', NULL ) ||

decode( sign( abs(&num) ), +1, to_char( to_date( abs(&num),'J'),'Jsp') ) from dual

Demonstrate simple encoding and decoding of messages

SELECT TRANSLATE(

'HELLO WORLD', -- Message to encode

'ABCDEFGHIJKLMNOPQRSTUVWXYZ ',

'1234567890!@#$%^&*()-=_+;,.') ENCODED_MESSAGE

FROM DUAL

SELECT TRANSLATE(

'85@@%._%*@4', -- Message to decode

'1234567890!@#$%^&*()-=_+;,.',

'ABCDEFGHIJKLMNOPQRSTUVWXYZ ') DECODED_MESSAGE

FROM DUAL

Friday, 27 April 2012

Pass application info through to the Oracle RDBMS

The following code tells the database what the application is up to:

begin

   dbms_application_info.set_client_info('BANCS application info');

   dbms_application_info.set_module('BANCS XYZ module', 'BANCS action name');

end;

-- Retrieve application info from the database:

select module, action, client_info

from   sys.v_$session where audsid = USERENV('SESSIONID')

select sql_text

from   sys.v_$sqlarea

where  module = 'BANCS XYZ module'

and  action = 'BANCS action name'

Display table and column comments

set pages 50000

set null 'No Comments'

tti 'Table Comments'

col comments format a29 wrap word

select * from user_tab_comments;

tti 'Column Comments'

col comments format a18 wrap word

break on table_name skip 1

select * from user_col_comments;

clear break

set null ''

set pages 23

Demonstrate default column values

drop table x

-- /

create table x (a char, b number default 99999, c date, d varchar2(6))

alter table x modify (c date default sysdate)

insert into x(a, d) values ('a', 'qwerty')

select * from x

--

-- Expected output:

--

--  A          B C           D

--  - ---------- ----------- ------

--  a      99999 25-APR-2001 qwerty

Select the Nth lowest value from a table

select level, min('col_name') from my_table

where level = '&n'

connect by prior ('col_name') < 'col_name')

group by level;

-- Example:

--

-- Given a table called emp with the following columns:

--   id   number

--   name varchar2(20)

--   sal  number

--

-- For the second lowest salary:

--

-- select level, min(sal) from emp

-- where level=2

-- connect by prior sal < sal

-- group by level

Select the Nth highest value from a table

select level, max('col_name') from my_table

where level = '&n'

connect by prior ('col_name') > 'col_name')

group by level;

-- Example :

--

-- Given a table called emp with the following columns:

--   id   number

--   name varchar2(20)

--   sal  number

--

-- For the second highest salary:

--

 select level, max(sal) from emp

 where level=2

 connect by prior sal > sal

 group by level

--

Thursday, 26 April 2012

Who am I, script

set termout off

store set store rep

set head off

set pause off

set termout on

select 'User: '|| user || ' on database ' || global_name,

       '  (term='||USERENV('TERMINAL')||

       ', audsid='||USERENV('SESSIONID')||')' as MYCONTEXT

from   global_name;

@store

set termout on

Display Database version, installed options and port string

set head off feed off pages 0 serveroutput on

col banner format a72 wrap

select banner

from   sys.v_$version;

select '   With the '||parameter||' option'

from   sys.v_$option

where  value = 'TRUE';

select '   The '||parameter||' option is not installed'

from   sys.v_$option

where  value <> 'TRUE';

begin

    dbms_output.put_line('Port String: '||dbms_utility.port_string);

end;

set head on feed on

Lookup Oracle error messages

set serveroutput on

set veri off feed off

prompt Lookup Oracle error messages:

prompt

prompt Please enter error numbers as negatives. E.g. -1

prompt

exec dbms_output.put_line('==> '||sqlerrm( &errno ) );

set veri on feed on

undef errno

Sample SQL matrix report

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

-- Sample output:

--

-- JOB           DEPT10     DEPT20     DEPT30     DEPT40

-- --------- ---------- ---------- ---------- ----------

-- ANALYST                    6000

-- CLERK           1300       1900        950

-- MANAGER         2450       2975       2850

-- PRESIDENT       5000

-- SALESMAN                              5600

--

26)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.

27)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.

SQL Interview questions -4

21)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;

22)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.

23)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;

24)Can one rename a column in a table?

From Oracle9i one can RENAME a column from a table. Look at this example:

ALTER TABLE tablename RENAME COLUMN oldcolumn TO newcolumn;

Other 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;

25)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".

Wednesday, 25 April 2012

SQL Interview questions -3

16)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).

Ravi Pachalla provided this solution:

        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);

Youssef Youssef provided this soluton: "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.

17)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 - provided by Ravi Pachalla

        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.

18)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;

19)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.

20)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

SQL Interview questions -2

11)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.

12)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

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
An 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 next month at 12 midnight	TRUNC(LAST_DAY(SYSDATE ) + 1)
First day of the current month	TRUNC(LAST_DAY(ADD_MONTHS(SYSDATE,-1))) + 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)

13)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;

14)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;

15)Can one retrieve only the Nth row from a table?

Rupak Mohan provided this solution to select the Nth row from a table:

        SELECT * FROM t1 a

        WHERE  n = (SELECT COUNT(rowid)

               FROM t1 b

               WHERE a.rowid >= b.rowid);

Shaik Khaleel provided this solution:

         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.

Ravi Pachalla provided these solutions:

        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);