//SQL语句之合并行列转换
//=====================================
有表rowtocol，其数据如下：
a b
1 1
1 2
1 3
2 1
2 2
3 1
如何转换成如下结果:
a b
1 1,2,3
2 1,2
3 1
创建一个合并的函数
create function f_rowtocol(@a int)
returns varchar(8000)
as
begin
declare @str varchar(8000)
set @str = ''
select @str = @str + ',' + cast(b as varchar) from rowtocol where a = @a
set @str = right(@str , len(@str) - 1)
return(@str)
End
go
调用自定义函数得到结果：
select distinct a ,dbo.f_rowtocol(a) from rowtocol
//=================================================================
//SQL语句之普通行列转换
//=================================================================
假设有张学生成绩表(tb_rowtocol)如下
Name Subject Result
张三 语文 73
张三 数学 83
张三 物理 93
李四 语文 74
李四 数学 84
李四 物理 94
想变成
姓名 语文 数学 物理
张三 73 83 93
李四 74 84 94
declare @sql varchar(4000)
set @sql = 'select Name as ' + '姓名'
select @sql = @sql + ' , sum(case Subject when '''+Subject+''' then Result end) ['+Subject+']'
from (select distinct Subject from rowtocol) as a
set @sql = @sql + ' from rowtocol group by name'
exec(@sql)
如果上述两表互相换一下：即
表名(cj)
姓名 语文 数学 物理
张三 73 83 93
李四 74 84 94
想变成
Name Subject Result
张三 语文 73
张三 数学 83
张三 物理 93
李四 语文 74
李四 数学 84
李四 物理 94
select 姓名 as Name,'语文' as Subject,语文 as Result from CJ union
select 姓名 as Name,'数学' as Subject,数学 as Result from CJ union
select 姓名 as Name,'物理' as Subject,物理 as Result from CJ
order by 姓名 desc
//===================================================
//行列转换加合计
//===================================================
例如有表A某些人每月消费数据如下：
id data month
001 11 1
001 12 2
001 13 3
001 14 4
001 15 5
001 16 6
001 17 7
001 18 8
001 19 9
001 110 10
001 111 11
001 112 12
002 21 1
002 22 2
002 23 3
002 24 4
002 25 5
002 26 6
002 27 7
002 28 8
002 29 9
002 210 10
002 211 11
002 212 12
要实现如下结果：
第一个：
SELECT id as '人员' ,
SUM(CASE month WHEN 1 THEN data ELSE 0 END) AS '1月' ,
SUM(CASE month WHEN 2 THEN data ELSE 0 END) AS '2月' ,
SUM(CASE month WHEN 3 THEN data ELSE 0 END) AS '3月' ,
SUM(CASE month WHEN 4 THEN data ELSE 0 END) AS '4月' ,
SUM(CASE month WHEN 5 THEN data ELSE 0 END) AS '5月' ,
SUM(CASE month WHEN 6 THEN data ELSE 0 END) AS '6月' ,
SUM(CASE month WHEN 7 THEN data ELSE 0 END) AS '7月' ,
SUM(CASE month WHEN 8 THEN data ELSE 0 END) AS '8月' ,
SUM(CASE month WHEN 9 THEN data ELSE 0 END) AS '9月' ,
SUM(CASE month WHEN 10 THEN data ELSE 0 END) AS '10月' ,
SUM(CASE month WHEN 11 THEN data ELSE 0 END) AS '11月' ,
SUM(CASE month WHEN 12 THEN data ELSE 0 END) AS '12月'
FROM a
GROUP BY ID
第二个：
SELECT id as '人员' ,
SUM(CASE month WHEN 1 THEN data ELSE 0 END) AS '1月' ,
SUM(CASE month WHEN 2 THEN data ELSE 0 END) AS '2月' ,
SUM(CASE month WHEN 3 THEN data ELSE 0 END) AS '3月' ,
SUM(CASE month WHEN 4 THEN data ELSE 0 END) AS '4月' ,
SUM(CASE month WHEN 5 THEN data ELSE 0 END) AS '5月' ,
SUM(CASE month WHEN 6 THEN data ELSE 0 END) AS '6月' ,
SUM(CASE month WHEN 7 THEN data ELSE 0 END) AS '7月' ,
SUM(CASE month WHEN 8 THEN data ELSE 0 END) AS '8月' ,
SUM(CASE month WHEN 9 THEN data ELSE 0 END) AS '9月' ,
SUM(CASE month WHEN 10 THEN data ELSE 0 END) AS '10月' ,
SUM(CASE month WHEN 11 THEN data ELSE 0 END) AS '11月' ,
SUM(CASE month WHEN 12 THEN data ELSE 0 END) AS '12月' ,
SUM(data) as '合计'
FROM a
GROUP BY ID
