---测试数据---
if object_id('[表1]') is not null drop table [表1]
go
create table [表1]([id] int,[姓名] varchar(1))
insert [表1]
select 1,'A' union all
select 2,'B' union all
select 3,'C' union all
select 3,'D' union all
select 3,'F' union all
select 4,'G' union all
select 5,'H' union all
select 5,'I' union all
select 6,'J'
---创建字符连接函数---
create function F_Str(@col1 varchar(20))
returns nvarchar(1000)
as
begin
declare @S nvarchar(1000)
select
@S=isnull(@S+'\','')+姓名
from
表1
where
Id=@col1
return @S
end---查询---
select
distinct
id,
dbo.f_str(id) as 姓名
from
[表1]---结果---
id 姓名
----------- --------------------
1 A
2 B
3 C\D\F
4 G
5 H\I
6 J(所影响的行数为 6 行)
if object_id('[表1]') is not null drop table [表1]
go
create table [表1]([id] int,[姓名] varchar(1))
insert [表1]
select 1,'A' union all
select 2,'B' union all
select 3,'C' union all
select 3,'D' union all
select 3,'F' union all
select 4,'G' union all
select 5,'H' union all
select 5,'I' union all
select 6,'J'
---创建字符连接函数---
create function F_Str(@col1 varchar(20))
returns nvarchar(1000)
as
begin
declare @S nvarchar(1000)
select
@S=isnull(@S+'\','')+姓名
from
表1
where
Id=@col1
return @S
end---查询---
select
distinct
id,
dbo.f_str(id) as 姓名
from
[表1]---结果---
id 姓名
----------- --------------------
1 A
2 B
3 C\D\F
4 G
5 H\I
6 J(所影响的行数为 6 行)
SQL code问题描述:
无论是在sql 2000,还是在 sql 2005 中,都没有提供字符串的聚合函数,
所以,当我们在处理下列要求时,会比较麻烦:
有表tb, 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-- 1. 创建处理函数
CREATE FUNCTION dbo.f_str(@id int)
RETURNS varchar(8000)
AS
BEGIN
DECLARE @r varchar(8000)
SET @r = ''
SELECT @r = @r + ',' + value
FROM tb
WHERE id=@id
RETURN STUFF(@r, 1, 1, '')
END
GO
-- 调用函数SELECt id, values=dbo.f_str(id)
FROM tb
GROUP BY id-- 2. 新的解决方法
-- 示例数据
DECLARE @t TABLE(id int, value varchar(10))
INSERT @t SELECT 1, 'aa'
UNION ALL SELECT 1, 'bb'
UNION ALL SELECT 2, 'aaa'
UNION ALL SELECT 2, 'bbb'
UNION ALL SELECT 2, 'ccc'-- 查询处理
SELECT *
FROM(
SELECT DISTINCT
id
FROM @t
)A
OUTER APPLY(
SELECT
[values]= STUFF(REPLACE(REPLACE(
(
SELECT value FROM @t N
WHERE id = A.id
FOR XML AUTO
), '<N value="', ','), '"/>', ''), 1, 1, '')
)N/*--结果
id values
----------- ----------------
1 aa,bb
2 aaa,bbb,ccc
(2 行受影响)
--*/
INSERT @t SELECT 1, 'aa'
UNION ALL SELECT 1, 'bb'
UNION ALL SELECT 2, 'aaa'
UNION ALL SELECT 2, 'bbb'
UNION ALL SELECT 2, 'ccc'-- 查询处理
SELECT *
FROM(
SELECT DISTINCT
id
FROM @t
)A
OUTER APPLY(
SELECT
[values]= STUFF(REPLACE(REPLACE(
(
SELECT value FROM @t N
WHERE id = A.id
FOR XML AUTO
), '<N value="', ','), '"/>', ''), 1, 1, '')
)N
这个是不是在2005中才能处理
go
create table [tb]([id] int,[姓名] varchar(10))
insert [tb] select 1,'A'
union all select 2,'B'
union all select 3,'C'
union all select 3,'D'
union all select 3,'F'
union all select 4,'G'
union all select 5,'H'
union all select 5,'I'
union all select 6,'J'
goselect distinct ID,stuff((select '\'+姓名 from tb where id=t.id for xml path('')),1,1,'') as 姓名 from tb t
/*
ID 姓名
----------- ------------
1 A
2 B
3 C\D\F
4 G
5 H\I
6 J(6 行受影响)
*/