我试图转动一张桌子并且遇到一些困难。我的表架构如下所示:
ID W_C W_P A_C A_P
3 257 251342 217 206078
4 443 109023 332 87437
6 17 9985 32 13515
我想要实现的目标是:
3 4 6
w_c 257 443 17
w_p 251342 109023 9985
a_c 217 332 87437
a_p 206078 87437 13515
我不需要显示w_c
,w_p
,a_c
或a_p
我只是将其用作参考点。
我认为可能有一种方法可以使用pivot / unpivot,但我对它们并不是很熟悉,而且我所读到的内容并没有帮助我。
我尝试使用CTE做一些事情,但我认为它过于复杂,只是不好的做法:
;with [3_Data] as (
Select
1 as [common_key3]
,max(Case when [id] = 3 then [w_c] else 0 end) as [3_Records]
From [example]
), [4_data] as (
Select
1 as [common_key4]
,max(Case when [id] = 4 then [w_c] else 0 end) as [4_Records]
From [example]
), [6_data] as (
Select
1 as [common_key6]
,max(Case when [id] = 6 then [w_c] else 0 end) as [6_Records]
From [example]
)
Select [3_Records], [4_Records], [6_Records]
From [3_Data]
inner join [4_data] on [common_key3] = [common_key4]
inner join [6_data] on [common_key3] = [common_key6]
Sql已经创建了表格http://sqlfiddle.com/#!3/02ef2/6
答案 0 :(得分:3)
您可以使用UNPIVOT
,然后使用PIVOT
来获得所需的结果(请参阅SQL Fiddle with Demo):
select col, [3], [4], [6]
from
(
select id, value, col
from
(
select id, w_c, w_p, a_c, a_p
from t
) x
unpivot
(
value
for col in (w_c, w_p, a_c, a_p)
)u
) x1
pivot
(
sum(value)
for id in ([3], [4], [6])
) p
在此查询中,您将首先UNPIVOT
,这样您就可以更轻松地访问要转换的数据。您可以对value
和col
字段使用任何名称,我只是为此示例选择了该名称。数据取消后,您可以应用PIVOT
来获取最终产品。