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ÇóÒ»SQLÓï¾ä - MS-SQL Server / »ù´¡Àà

ÓбíÈçÏÂ
a b c
001 ÊÕÈë 5000
001 ·ÑÓÃ 2000
001 ºÏ¼Æ 3000
002 ÊÕÈë 6000
002 ·ÑÓÃ 3000
002 ºÏ¼Æ 3000
003 ÊÕÈë 4000
003 ·ÑÓÃ 1000
003 ºÏ¼Æ 2000
...
ºÏ¼Æ=ÊÕÈë-·ÑÓÃ
ÇóÒ»Óï¾ä ²éÕÒ³öµ±b=ÊÕÈëʱc×ֶβ»ÕýÈ·µÄ¼Ç¼


select ²»ÕýÈ·¼Ç¼=x.a from ta x join ta y on x.ºÏ¼Æ<>y.ÊÕÈë-y.·ÑÓà where x.b='ÊÕÈë'

SQL code:

select a,sum(case b when 'ÊÕÈë' then c else -c end)
from your_table
group by a
having sum(case b when 'ÊÕÈë' then c else -c end) <> 0



SQL code:

use PracticeDB
go
if exists (select 1 from sysobjects where name='tb_a')
drop table tb_a
go
create table tb_a (a varchar(10), b varchar(10),c numeric(10))
go
insert into tb_a
select '001', 'ÊÕÈë' ,5000 union all
select '001', '·ÑÓÃ' ,2000 union all
select '001', 'ºÏ¼Æ' ,3000 union all
select '002', 'ÊÕÈë' ,6000 union all
select '002', '·ÑÓÃ' ,3000 union all
select '002', 'ºÏ¼Æ' ,3000 union all
select '003', 'ÊÕÈë' ,4000 union all
select '003', '·ÑÓÃ' ,1000 union all
select '003', 'ºÏ¼Æ' ,2000
ºÏ¼Æ=ÊÕÈë-·ÑÓÃ
ÇóÒ»Óï¾ä ²éÕÒ³öµ±b=ÊÕÈëʱc×ֶβ»ÕýÈ·µÄ¼Ç¼


select * from tb_a

;with t
as
(
select a,sum(case b when 'ÊÕÈë' then c else 0 end) [ÊÕÈë],


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