Use PERCENT_RANK Function : PERCENT_RANK « Analytical Functions « Oracle PL/SQL Tutorial





The syntax for PERCENT_RANK,

PERCENT_RANK() OVER ([PARTITION clause] ORDER clause)

The PARTITION clause is optional.

SQL>
SQL> -- create demo table
SQL> create table Employee(
  2    ID                 VARCHAR2(4 BYTE)         NOT NULL,
  3    First_Name         VARCHAR2(10 BYTE),
  4    Last_Name          VARCHAR2(10 BYTE),
  5    Start_Date         DATE,
  6    End_Date           DATE,
  7    Salary             Number(8,2),
  8    City               VARCHAR2(10 BYTE),
  9    Description        VARCHAR2(15 BYTE)
 10  )
 11  /

Table created.

SQL>
SQL> -- prepare data
SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary,  City,       Description)
  2               values ('01','Jason',    'Martin',  to_date('19960725','YYYYMMDD'), to_date('20060725','YYYYMMDD'), 1234.56, 'Toronto',  'Programmer')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary,  City,       Description)
  2                values('02','Alison',   'Mathews', to_date('19760321','YYYYMMDD'), to_date('19860221','YYYYMMDD'), 6661.78, 'Vancouver','Tester')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary,  City,       Description)
  2                values('03','James',    'Smith',   to_date('19781212','YYYYMMDD'), to_date('19900315','YYYYMMDD'), 6544.78, 'Vancouver','Tester')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary,  City,       Description)
  2                values('04','Celia',    'Rice',    to_date('19821024','YYYYMMDD'), to_date('19990421','YYYYMMDD'), 2344.78, 'Vancouver','Manager')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary,  City,       Description)
  2                values('05','Robert',   'Black',   to_date('19840115','YYYYMMDD'), to_date('19980808','YYYYMMDD'), 2334.78, 'Vancouver','Tester')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary, City,        Description)
  2                values('06','Linda',    'Green',   to_date('19870730','YYYYMMDD'), to_date('19960104','YYYYMMDD'), 4322.78,'New York',  'Tester')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary, City,        Description)
  2                values('07','David',    'Larry',   to_date('19901231','YYYYMMDD'), to_date('19980212','YYYYMMDD'), 7897.78,'New York',  'Manager')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary, City,        Description)
  2                values('08','James',    'Cat',     to_date('19960917','YYYYMMDD'), to_date('20020415','YYYYMMDD'), 1232.78,'Vancouver', 'Tester')
  3  /

1 row created.

SQL>
SQL>
SQL>
SQL> -- display data in the table
SQL> select * from Employee
  2  /

ID   FIRST_NAME LAST_NAME  START_DAT END_DATE      SALARY CITY       DESCRIPTION
---- ---------- ---------- --------- --------- ---------- ---------- ---------------
01   Jason      Martin     25-JUL-96 25-JUL-06    1234.56 Toronto    Programmer
02   Alison     Mathews    21-MAR-76 21-FEB-86    6661.78 Vancouver  Tester
03   James      Smith      12-DEC-78 15-MAR-90    6544.78 Vancouver  Tester
04   Celia      Rice       24-OCT-82 21-APR-99    2344.78 Vancouver  Manager
05   Robert     Black      15-JAN-84 08-AUG-98    2334.78 Vancouver  Tester
06   Linda      Green      30-JUL-87 04-JAN-96    4322.78 New York   Tester
07   David      Larry      31-DEC-90 12-FEB-98    7897.78 New York   Manager
08   James      Cat        17-SEP-96 15-APR-02    1232.78 Vancouver  Tester

8 rows selected.

SQL>
SQL>
SQL> SELECT first_name, salary,
  2    RANK() OVER(ORDER BY salary) RANK,
  3    PERCENT_RANK() OVER(ORDER BY salary) PR,
  4    CUME_DIST() OVER(ORDER BY salary) CD
  5  FROM employee
  6  ORDER BY salary
  7  /

FIRST_NAME     SALARY       RANK         PR         CD
---------- ---------- ---------- ---------- ----------
James         1232.78          1          0       .125
Jason         1234.56          2 .142857143        .25
Robert        2334.78          3 .285714286       .375
Celia         2344.78          4 .428571429         .5
Linda         4322.78          5 .571428571       .625
James         6544.78          6 .714285714        .75
Alison        6661.78          7 .857142857       .875
David         7897.78          8          1          1

8 rows selected.

SQL>
SQL>
SQL>
SQL> -- clean the table
SQL> drop table Employee
  2  /

Table dropped.










16.6.PERCENT_RANK
16.6.1.PERCENT_RANK
16.6.2.Use PERCENT_RANK Function
16.6.3.PERCENT_RANK() calculates the percent rank of a value relative to a group of values.
16.6.4.Using the Hypothetical Rank and Distribution Functions