---测试数据---
if object_id('[tb]') is not null drop table [tb]
go
create table [tb]([id] int,[values] varchar(1))
insert [tb]
select 1,'x' union all
select 1,'y' union all
select 2,'a' union all
select 2,'b' union all
select 2,'c'
---创建字符连接函数---
create function F_Str(@col1 int)
returns nvarchar(100)
as
begin
declare @S nvarchar(100)
select
@S=isnull(@S+'','')+[values]
from
tb
where
Id=@col1
return @S
end---查询---
select
distinct
id,
dbo.f_str(id) as [values]
from
[tb]
---结果---
id values
----------- -------------------
1 xy
2 abc(所影响的行数为 2 行)
if object_id('[tb]') is not null drop table [tb]
go
create table [tb]([id] int,[values] varchar(1))
insert [tb]
select 1,'x' union all
select 1,'y' union all
select 2,'a' union all
select 2,'b' union all
select 2,'c'
---创建字符连接函数---
create function F_Str(@col1 int)
returns nvarchar(100)
as
begin
declare @S nvarchar(100)
select
@S=isnull(@S+'','')+[values]
from
tb
where
Id=@col1
return @S
end---查询---
select
distinct
id,
dbo.f_str(id) as [values]
from
[tb]
---结果---
id values
----------- -------------------
1 xy
2 abc(所影响的行数为 2 行)
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)
--*/--各种字符串分函数--3.3.1 使用游标法进行字符串合并处理的示例。
--处理的数据
CREATE TABLE tb(col1 varchar(10),col2 int)
INSERT tb SELECT 'a',1
UNION ALL SELECT 'a',2
UNION ALL SELECT 'b',1
UNION ALL SELECT 'b',2
UNION ALL SELECT 'b',3--合并处理
--定义结果集表变量
DECLARE @t TABLE(col1 varchar(10),col2 varchar(100))--定义游标并进行合并处理
DECLARE tb CURSOR LOCAL
FOR
SELECT col1,col2 FROM tb ORDER BY col1,col2
DECLARE @col1_old varchar(10),@col1 varchar(10),@col2 int,@s varchar(100)
OPEN tb
FETCH tb INTO @col1,@col2
SELECT @col1_old=@col1,@s=''
WHILE @@FETCH_STATUS=0
BEGIN
IF @col1=@col1_old
SELECT @s=@s+','+CAST(@col2 as varchar)
ELSE
BEGIN
INSERT @t VALUES(@col1_old,STUFF(@s,1,1,''))
SELECT @s=','+CAST(@col2 as varchar),@col1_old=@col1
END
FETCH tb INTO @col1,@col2
END
INSERT @t VALUES(@col1_old,STUFF(@s,1,1,''))
CLOSE tb
DEALLOCATE tb
--显示结果并删除测试数据
SELECT * FROM @t
DROP TABLE tb
/*--结果
col1 col2
---------- -----------
a 1,2
b 1,2,3
--*/
GO
if object_id('[LI]') is not null drop table [LI]
go
create table [LI]([id] int,[values] varchar(1))
insert [LI]
select 1,'x' union all
select 1,'y' union all
select 2,'a' union all
select 2,'b' union all
select 2,'c'
SELECT *
FROM(
SELECT DISTINCT
id
FROM LI
)A
OUTER APPLY(
SELECT
[values]= REPLACE(REPLACE(
(
SELECT [values] FROM LI N
WHERE id = A.id
FOR XML AUTO
), '<N values="', ''), '"/>', '')
)N/*
id values
----------- ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
1 xy
2 abc
*/
@S=isnull(@S+'','')+[values]
这句话是什么意思呢?
create function F_Str(@col1 int)
还有这句话谢谢