SQL Server/T-SQL Tutorial/Aggregate Functions/RANK
RANK() OVER
<source lang="sql">
5> create table department( 6> dept_name char(20) not null, 7> emp_cnt int not null, 8> budget float, 9> date_month datetime); 10> GO 1> 2> insert into department values("Research", 5, 50000, "01.01.2002"); 3> insert into department values("Research", 10, 70000, "01.02.2002"); 4> insert into department values("Research", 5, 65000, "01.07.2002"); 5> insert into department values("Accounting", 5, 10000, "01.07.2002"); 6> insert into department values("Accounting", 10, 40000, "01.02.2002"); 7> insert into department values("Accounting", 6, 30000, "01.01.2002"); 8> insert into department values("Accounting", 6, 40000, "01.02.2003"); 9> insert into department values("Marketing", 6, 10000, "01.01.2003"); 10> insert into department values("Marketing", 10, 40000, "01.02.2003"); 11> insert into department values("Marketing", 3, 30000, "01.07.2003"); 12> insert into department values("Marketing", 5, 40000, "01.01.2003"); 13> GO (1 rows affected) (1 rows affected) (1 rows affected) (1 rows affected) (1 rows affected) (1 rows affected) (1 rows affected) (1 rows affected) (1 rows affected) (1 rows affected) (1 rows affected) 1> 2> SELECT RANK() OVER(ORDER BY budget DESC) AS rank_budget,dept_name, emp_cnt, budget 3> FROM department 4> WHERE budget <= 50000; 5> GO rank_budget dept_name emp_cnt budget
-------------------- ----------- ------------------------
1 Research 5 50000
2 Accounting 10 40000
2 Accounting 6 40000
2 Marketing 10 40000
2 Marketing 5 40000
6 Marketing 3 30000
6 Accounting 6 30000
8 Accounting 5 10000
8 Marketing 6 10000
(9 rows affected) 1> drop table department; 2> GO 1> 2> 3></source>
Returning Rows by Rank Without Gaps
<source lang="sql">
4> 5> CREATE TABLE employee( 6> id INTEGER NOT NULL PRIMARY KEY, 7> first_name VARCHAR(10), 8> last_name VARCHAR(10), 9> salary DECIMAL(10,2), 10> start_Date DATETIME, 11> region VARCHAR(10), 12> city VARCHAR(20), 13> managerid INTEGER 14> ); 15> GO 1> INSERT INTO employee VALUES (1, "Jason" , "Martin", 5890,"2005-03-22","North","Vancouver",3); 2> GO (1 rows affected) 1> INSERT INTO employee VALUES (2, "Alison", "Mathews",4789,"2003-07-21","South","Utown",4); 2> GO (1 rows affected) 1> INSERT INTO employee VALUES (3, "James" , "Smith", 6678,"2001-12-01","North","Paris",5); 2> GO (1 rows affected) 1> INSERT INTO employee VALUES (4, "Celia" , "Rice", 5567,"2006-03-03","South","London",6); 2> GO (1 rows affected) 1> INSERT INTO employee VALUES (5, "Robert", "Black", 4467,"2004-07-02","East","Newton",7); 2> GO (1 rows affected) 1> INSERT INTO employee VALUES (6, "Linda" , "Green" , 6456,"2002-05-19","East","Calgary",8); 2> GO (1 rows affected) 1> INSERT INTO employee VALUES (7, "David" , "Larry", 5345,"2008-03-18","West","New York",9); 2> GO (1 rows affected) 1> INSERT INTO employee VALUES (8, "James" , "Cat", 4234,"2007-07-17","West","Regina",9); 2> GO (1 rows affected) 1> INSERT INTO employee VALUES (9, "Joan" , "Act", 6123,"2001-04-16","North","Toronto",10); 2> GO (1 rows affected) 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> 4> SELECT ID,Salary, 5> DENSE_RANK() OVER (ORDER BY Salary DESC) as DENSE_RANK 6> FROM Employee 7> GO ID Salary DENSE_RANK
------------ --------------------
3 6678.00 1
6 6456.00 2
9 6123.00 3
1 5890.00 4
4 5567.00 5
7 5345.00 6
2 4789.00 7
5 4467.00 8
8 4234.00 9
(9 rows affected) 1> 2> drop table employee; 3> GO</source>
The RANK() function preserves the ordinal position of the row in the list.
<source lang="sql">
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)</source>
