The ranking as opposed to a row-numbering problem (the problem of ties) : Rank « Analytical Functions « Oracle PL / SQL






The ranking as opposed to a row-numbering problem (the problem of ties)

 

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'), 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('03','James',    'Smith',   to_date('19781212','YYYYMMDD'), to_date('19900315','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('04','Celia',    'Rice',    to_date('19821024','YYYYMMDD'), to_date('19990421','YYYYMMDD'), 2334.78, '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'), 2334.78,'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'), 2334.78,'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'), 2334.78,'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    2334.78 Vancouver  Tester
03   James      Smith      12-DEC-78 15-MAR-90    2334.78 Vancouver  Tester
04   Celia      Rice       24-OCT-82 21-APR-99    2334.78 Vancouver  Manager
05   Robert     Black      15-JAN-84 08-AUG-98    2334.78 Vancouver  Tester
06   Linda      Green      30-JUL-87 04-JAN-96    2334.78 New York   Tester
07   David      Larry      31-DEC-90 12-FEB-98    2334.78 New York   Manager
08   James      Cat        17-SEP-96 15-APR-02    2334.78 Vancouver  Tester

8 rows selected.

SQL>
SQL> -- the ranking as opposed to a row-numbering problem (the problem of ties),
SQL>
SQL> SELECT id, first_name, salary,
  2    RANK() OVER(ORDER BY salary desc) toprank
  3  FROM employee;

ID   FIRST_NAME     SALARY    TOPRANK
---- ---------- ---------- ----------
04   Celia         2334.78          1
02   Alison        2334.78          1
03   James         2334.78          1
08   James         2334.78          1
05   Robert        2334.78          1
06   Linda         2334.78          1
07   David         2334.78          1
01   Jason         1234.56          8

8 rows selected.

SQL>
SQL>
SQL>
SQL>
SQL> -- clean the table
SQL> drop table Employee;

Table dropped.

SQL>
SQL>
SQL>

           
         
  








Related examples in the same category

1.RANK(): rank items, leaves a gap in the sequence when there is a tie
2.The WHERE is applied before the RANK()
3.Rank with order by clause
4.When RANK() is added to the statement, the RANK function is applied last, just before the ordering
5.RANK() with NULLS LAST demo
6.RANK() with NULLS FIRST
7.Rank(): If the statement requests another ordering, another sort may result
8.Rank() with null values
9.Rank() over partition
10.Browse Products with ranking function
11.rank() over (partition by deptno order by sal desc )