SQL Server/T-SQL/Aggregate Functions/Aggregate Function Basics
Null value is eliminated by an aggregate or other SET operation
<source lang="sql">
1> create table employee( 2> ID int, 3> name nvarchar (10), 4> salary int, 5> start_date datetime, 6> city nvarchar (10), 7> region char (1)) 8> GO 1> 2> insert into employee (ID, name, salary, start_date, city, region) 3> values (1, "Jason", 40420, "02/01/94", "New York", "W") 4> GO (1 rows affected) 1> insert into employee (ID, name, salary, start_date, city, region) 2> values (2, "Robert",14420, "01/02/95", "Vancouver","N") 3> GO (1 rows affected) 1> insert into employee (ID, name, salary, start_date, city, region) 2> values (3, "Celia", 24020, "12/03/96", "Toronto", "W") 3> GO (1 rows affected) 1> insert into employee (ID, name, salary, start_date, city, region) 2> values (4, "Linda", 40620, "11/04/97", "New York", "N") 3> GO (1 rows affected) 1> insert into employee (ID, name, salary, start_date, city, region) 2> values (5, "David", 80026, "10/05/98", "Vancouver","W") 3> GO (1 rows affected) 1> insert into employee (ID, name, salary, start_date, city, region) 2> values (6, "James", 70060, "09/06/99", "Toronto", "N") 3> GO (1 rows affected) 1> insert into employee (ID, name, salary, start_date, city, region) 2> values (7, "Alison",90620, "08/07/00", "New York", "W") 3> GO (1 rows affected) 1> insert into employee (ID, name, salary, start_date, city, region) 2> values (8, "Chris", 26020, "07/08/01", "Vancouver","N") 3> GO (1 rows affected) 1> insert into employee (ID, name, salary, start_date, city, region) 2> values (9, "Mary", 60020, "06/09/02", "Toronto", "W") 3> GO (1 rows affected) 1> 2> select * from employee 3> GO ID name salary start_date city region
---------- ----------- ----------------------- ---------- ------
1 Jason 40420 1994-02-01 00:00:00.000 New York W
2 Robert 14420 1995-01-02 00:00:00.000 Vancouver N
3 Celia 24020 1996-12-03 00:00:00.000 Toronto W
4 Linda 40620 1997-11-04 00:00:00.000 New York N
5 David 80026 1998-10-05 00:00:00.000 Vancouver W
6 James 70060 1999-09-06 00:00:00.000 Toronto N
7 Alison 90620 2000-08-07 00:00:00.000 New York W
8 Chris 26020 2001-07-08 00:00:00.000 Vancouver N
9 Mary 60020 2002-06-09 00:00:00.000 Toronto W
(9 rows affected) 1> 2> SELECT COUNT(Name) 3> FROM Employee 4> GO
9
(1 rows affected) 1> 2> update Employee set name = null; 3> GO (9 rows affected) 1> 2> SELECT COUNT(Name) 3> FROM Employee 4> GO
0
Warning: Null value is eliminated by an aggregate or other SET operation. 1> 2> 3> 4> 5> 6> drop table employee 7> GO 1>
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Use sum function in a table join
<source lang="sql">
1> create table employee( 2> ID int, 3> name nvarchar (10), 4> salary int ) 5> GO 1> 2> create table job( 3> ID int, 4> title nvarchar (10), 5> averageSalary int) 6> GO 1> 2> 3> insert into employee (ID, name, salary) values (1, "Jason", 1234) 4> GO 1> insert into employee (ID, name, salary) values (2, "Robert", 4321) 2> GO 1> insert into employee (ID, name, salary) values (3, "Celia", 5432) 2> GO 1> insert into employee (ID, name, salary) values (4, "Linda", 3456) 2> GO 1> insert into employee (ID, name, salary) values (5, "David", 7654) 2> GO 1> insert into employee (ID, name, salary) values (6, "James", 4567) 2> GO 1> insert into employee (ID, name, salary) values (7, "Alison", 8744) 2> GO 1> insert into employee (ID, name, salary) values (8, "Chris", 9875) 2> GO 1> insert into employee (ID, name, salary) values (9, "Mary", 2345) 2> GO 1> 2> insert into job(ID, title, averageSalary) values(1,"Developer",3000) 3> GO 1> insert into job(ID, title, averageSalary) values(2,"Tester", 4000) 2> GO 1> insert into job(ID, title, averageSalary) values(3,"Designer", 5000) 2> GO 1> insert into job(ID, title, averageSalary) values(4,"Programmer", 6000) 2> GO 1> 2> 3> select * from employee; 4> GO ID name salary
---------- -----------
1 Jason 1234
2 Robert 4321
3 Celia 5432
4 Linda 3456
5 David 7654
6 James 4567
7 Alison 8744
8 Chris 9875
9 Mary 2345
1> select * from job; 2> GO ID title averageSalary
---------- -------------
1 Developer 3000
2 Tester 4000
3 Designer 5000
4 Programmer 6000
1> 2> SELECT sum(e.salary), j.title 3> FROM Employee e CROSS JOIN job j 4> group by j.title 5> GO
title
----------
47628 Designer
47628 Developer
47628 Programmer
47628 Tester
1> 2> 3> drop table employee; 4> drop table job; 5> GO 1> 2>
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