CREATE TABLE [dbo].[abc](
[a] [int] NULL,
[b] [int] NULL,
[c] [int] NULL
) ON [PRIMARY]insert into abc values(1,1,10)
insert into abc values(1,2,20)
insert into abc values(1,3,30)
insert into abc values(1,1,30)insert into abc values(2,1,30)
insert into abc values(2,1,20)
insert into abc values(2,3,10)insert into abc values(3,1,50)
insert into abc values(3,2,80)
....以下是执行的结果
m n o p
3 130 2,1 80,50
1 90 1,3,2 40,30,20
2 60 1,3 50,10
....说明一下,先对a进行分组,合计c 按c的合计值进行排序同时打印出合计值,然后
对b进行分组,合计c 按c的合计值进行排序用,号进行分隔打印,想了很久写不出来,求助大家,谢谢.
[a] [int] NULL,
[b] [int] NULL,
[c] [int] NULL
) ON [PRIMARY]insert into abc values(1,1,10)
insert into abc values(1,2,20)
insert into abc values(1,3,30)
insert into abc values(1,1,30)insert into abc values(2,1,30)
insert into abc values(2,1,20)
insert into abc values(2,3,10)insert into abc values(3,1,50)
insert into abc values(3,2,80)
....以下是执行的结果
m n o p
3 130 2,1 80,50
1 90 1,3,2 40,30,20
2 60 1,3 50,10
....说明一下,先对a进行分组,合计c 按c的合计值进行排序同时打印出合计值,然后
对b进行分组,合计c 按c的合计值进行排序用,号进行分隔打印,想了很久写不出来,求助大家,谢谢.
--*******************************************************************************************
表结构,数据如下:
id value
----- ------
1 aa
1 bb
2 aaa
2 bbb
2 ccc 需要得到结果:
id values
------ -----------
1 aa,bb
2 aaa,bbb,ccc
即:group by id, 求 value 的和(字符串相加) 1. 旧的解决方法(在sql server 2000中只能用函数解决。)
--=============================================================================
create table tb(id int, value varchar(10))
insert into tb values(1, 'aa')
insert into tb values(1, 'bb')
insert into tb values(2, 'aaa')
insert into tb values(2, 'bbb')
insert into tb values(2, 'ccc')
go
--1. 创建处理函数
CREATE FUNCTION dbo.f_strUnite(@id int)
RETURNS varchar(8000)
AS
BEGIN
DECLARE @str varchar(8000)
SET @str = ''
SELECT @str = @str + ',' + value FROM tb WHERE id=@id
RETURN STUFF(@str, 1, 1, '')
END
GO
-- 调用函数
SELECt id, value = dbo.f_strUnite(id) FROM tb GROUP BY id
drop table tb
drop function dbo.f_strUnite
go
/*
id value
----------- -----------
1 aa,bb
2 aaa,bbb,ccc
(所影响的行数为 2 行)
*/
--===================================================================================
2. 新的解决方法(在sql server 2005中用OUTER APPLY等解决。)
create table tb(id int, value varchar(10))
insert into tb values(1, 'aa')
insert into tb values(1, 'bb')
insert into tb values(2, 'aaa')
insert into tb values(2, 'bbb')
insert into tb values(2, 'ccc')
go
-- 查询处理
SELECT * FROM(SELECT DISTINCT id FROM tb)A OUTER APPLY(
SELECT [values]= STUFF(REPLACE(REPLACE(
(
SELECT value FROM tb N
WHERE id = A.id
FOR XML AUTO
), ' <N value="', ','), '"/>', ''), 1, 1, '')
)N
drop table tb /*
id values
----------- -----------
1 aa,bb
2 aaa,bbb,ccc (2 行受影响)
*/ --SQL2005中的方法2
create table tb(id int, value varchar(10))
insert into tb values(1, 'aa')
insert into tb values(1, 'bb')
insert into tb values(2, 'aaa')
insert into tb values(2, 'bbb')
insert into tb values(2, 'ccc')
go select id, [values]=stuff((select ','+[value] from tb t where id=tb.id for xml path('')), 1, 1, '')
from tb
group by id /*
id values
----------- --------------------
1 aa,bb
2 aaa,bbb,ccc (2 row(s) affected) */
select a,b,sum(c) as c
from abc
group by a,b
)
select a as m
,sum(c) as n
,o = =stuff((select ','+rtrim([b]) from cte t where a=a.a order by c desc for xml path('')), 1, 1, '')
,p = =stuff((select ','+rtrim([c]) from cte t where a=a.a order by c desc for xml path('')), 1, 1, '')
from cte a
group by a
[a] [int] NULL,
[b] [int] NULL,
[c] [int] NULL
) ON [PRIMARY]insert into abc values(1,1,10)
insert into abc values(1,2,20)
insert into abc values(1,3,30)
insert into abc values(1,1,30)insert into abc values(2,1,30)
insert into abc values(2,1,20)
insert into abc values(2,3,10)insert into abc values(3,1,50)
insert into abc values(3,2,80)select a,SUM(c) n,o=stuff((select ','+rtrim(b) from abc
where a=t.a group by b order by SUM(c) desc for xml path('')),1,1,''),
p=stuff((select ','+rtrim(sum(c)) from abc
where a=t.a group by b order by SUM(c) desc for xml path('')),1,1,'')
from abc t
group by a order by 2 desc/*
a n o p
----------- ----------- -------- ---------
3 130 2,1 80,50
1 90 1,3,2 40,30,20
2 60 1,3 50,10
a,sum(c) as n,
o=stuff((select ','+rtrim(b) from abc where a=t.a group by b order by SUM(c) desc for xml path('')),1,1,''),
p=stuff((select ','+rtrim(sum(c)) from abc where a=t.a group by b order by SUM(c) desc for xml path('')),1,1,'')
from
abc t
group by
a
order by
sum(c) desc