The RANK() function preserves the ordinal position of the row in the list. : RANK « Aggregate Functions « SQL Server / T-SQL Tutorial






RANK function increments its value for each row in the set.
The syntax for RANK is as follows:
RANK ( )   OVER ( [ < partition_by_clause > ] < order_by_clause > )
If rows with tied values exist, they will receive the same rank value

For each duplicate value, the RANK() function skips the subsequent value so that the next non-duplicate value remains in its rightful position.

7>
8> CREATE TABLE employee(
9>    id          INTEGER NOT NULL PRIMARY KEY,
10>    first_name  VARCHAR(10),
11>    last_name   VARCHAR(10),
12>    salary      DECIMAL(10,2),
13>    start_Date  DATETIME,
14>    region      VARCHAR(10),
15>    city        VARCHAR(20),
16>    managerid   INTEGER
17> );
18> GO
1> INSERT INTO employee VALUES (1, 'Jason' ,  'Martin', 5890,'2005-03-22','North','Vancouver',3);
2> GO
1> INSERT INTO employee VALUES (2, 'Alison',  'Mathews',4789,'2003-07-21','South','Utown',4);
2> GO
1> INSERT INTO employee VALUES (3, 'James' ,  'Smith',  6678,'2001-12-01','North','Paris',5);
2> GO
1> INSERT INTO employee VALUES (4, 'Celia' ,  'Rice',   5567,'2006-03-03','South','London',6);
2> GO
1> INSERT INTO employee VALUES (5, 'Robert',  'Black',  4467,'2004-07-02','East','Newton',7);
2> GO
1> INSERT INTO employee VALUES (6, 'Linda' ,  'Green' , 6456,'2002-05-19','East','Calgary',8);
2> GO
1> INSERT INTO employee VALUES (7, 'David' ,  'Larry',  5345,'2008-03-18','West','New York',9);
2> GO
1> INSERT INTO employee VALUES (8, 'James' ,  'Cat',    4234,'2007-07-17','West','Regina',9);
2> GO
1> INSERT INTO employee VALUES (9, 'Joan'  ,  'Act',    6123,'2001-04-16','North','Toronto',10);
2> GO
1>
2> select * from employee;
3> GO
id          first_name last_name  salary       start_Date              region     city                 managerid
----------- ---------- ---------- ------------ ----------------------- ---------- -------------------- -----------
          1 Jason      Martin          5890.00 2005-03-22 00:00:00.000 North      Vancouver                      3
          2 Alison     Mathews         4789.00 2003-07-21 00:00:00.000 South      Utown                          4
          3 James      Smith           6678.00 2001-12-01 00:00:00.000 North      Paris                          5
          4 Celia      Rice            5567.00 2006-03-03 00:00:00.000 South      London                         6
          5 Robert     Black           4467.00 2004-07-02 00:00:00.000 East       Newton                         7
          6 Linda      Green           6456.00 2002-05-19 00:00:00.000 East       Calgary                        8
          7 David      Larry           5345.00 2008-03-18 00:00:00.000 West       New York                       9
          8 James      Cat             4234.00 2007-07-17 00:00:00.000 West       Regina                         9
          9 Joan       Act             6123.00 2001-04-16 00:00:00.000 North      Toronto                       10

(9 rows affected)
1>
2>
3> SELECT
4>       ID
5>     , First_Name
6>     , RANK() Over (ORDER BY First_Name) As Rank
7> FROM Employee
8> ORDER BY First_Name
9>
10>
11> drop table employee;
12> GO
ID          First_Name Rank
----------- ---------- --------------------
          2 Alison                        1
          4 Celia                         2
          7 David                         3
          8 James                         4
          3 James                         4
          1 Jason                         6
          9 Joan                          7
          6 Linda                         8
          5 Robert                        9

(9 rows affected)








9.6.RANK
9.6.1.The RANK() function preserves the ordinal position of the row in the list.
9.6.2.RANK() OVER
9.6.3.Returning Rows by Rank Without Gaps