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if object_id('[tb]') is not null drop table [tb]
go
create table [tb]([username] varchar(1),[accessories] varchar(20))
insert [tb]
select 'A','附件1,附件2,,附件3,,' union all
select 'A','附件4,,附件5,附件6,,' union all
select 'A','附件7,附件8,,附件9,,' union all
select 'B','附件1,,附件2,附件3,,' union all
select 'B','附件4,,附件5,附件6,,' union all
select 'B','附件7,附件8,,附件9,,'
go
--select * from [tb]
select username,accessories=left(accessories,len(accessories)-1)
from
(
select username,accessories=(select replace(accessories,',,',',') from tb where username=t.username for xml path(''))
from tb t
group by username
) tt
/*
username accessories
-------- ---------------------------------------------------------
A 附件1,附件2,附件3,附件4,附件5,附件6,附件7,附件8,附件9
B 附件1,附件2,附件3,附件4,附件5,附件6,附件7,附件8,附件9
(2 行受影响)
*/
-- 测试数据
IF OBJECT_ID('LI') IS NOT NULL
DROP TABLE LI
CREATE TABLE LI (username varchar(10), accessories varchar(50))
INSERT LI SELECT 'A', '附件1,附件2,附件3,'
UNION ALL SELECT 'A', '附件4,附件5,附件6,'
UNION ALL SELECT 'A', '附件7,附件8,附件9,'
UNION ALL SELECT 'B', '附件1,附件2,附件3,'
UNION ALL SELECT 'B', '附件4,附件5,附件6,'
UNION ALL SELECT 'B', '附件7,附件8,附件9,'
--楼主要求1(合并数据)
SELECT *
FROM(
SELECT DISTINCT username
FROM LI
)A
OUTER APPLY(
SELECT
[accessoriesList]= STUFF(REPLACE(REPLACE(
(
SELECT [accessories] FROM LI N
WHERE [username] = A.username
FOR XML AUTO
), '<N accessories="', ','), '"/>', ''), 1, 1, '')
)N
--楼主要求2(计算数组的个数)
select LO.username, COUNT(accessories) AS accesscount
FROM
(
Select
a.username,b.accessories
From
(select username,accessories=convert(xml,'<root><v>'+replace(left(accessories,len(accessories)-1),',','</v><v>')+'</v></root>') from LI)a
outer apply
(select accessories=C.v.value('.','nvarchar(100)') from a.accessories.nodes('/root/v')C(v) )b
) LO
GROUP BY LO.USERNAME
--删除测试数据
DROP TABLE LI
/*
--结果1
username accessoriesList
A 附件1,附件2,附件3,附件4,附件5,附件6,,附件7,附件8,附件9,
B 附件1,附件2,附件3,附件4,附件5,附件6,,附件7,附件8,附件9,
--结果1
username accesscount
A 9
B 9
*/
-- 测试数据
IF OBJECT_ID('LI') IS NOT NULL
DROP TABLE LI
CREATE TABLE LI (username varchar(10), accessories varchar(50))
INSERT LI SELECT 'A', '附件1,附件2,附件3,'
UNION ALL SELECT 'A', '附件4,附件5,附件6,'
UNION ALL SELECT 'A', '附件7,附件8,附件9,'
UNION ALL SELECT 'B', '附件1,附件2,附件3,'
UNION ALL SELECT 'B', '附件4,附件5,附件6,'
UNION ALL SELECT 'B', '附件7,附件8,附件9,'
-- 查询处理
SELECT *
FROM(
SELECT DISTINCT username
FROM LI
)A
OUTER APPLY(
SELECT
[accessoriesList]= STUFF(REPLACE(REPLACE(
(
SELECT [accessories] FROM LI N
WHERE [username] = A.username
FOR XML AUTO
), '<N accessories="', ','), '"/>', ''), 1, 1, '')
)N
--******************************************************************************************
-- 合并列值
--*******************************************************************************************
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)
*/
drop table tb