我有东西
通过运行脚本获得
declare @t table(id int identity, bucket varchar(200), startinventory int, nocontact int)
INSERT INTO @t
SELECT 'bucket1',1234,72500 UNION ALL
SELECT 'bucket2',6784,60500 UNION ALL
SELECT 'bucket3',678,52000 UNION ALL
SELECT 'bucket4',234,45000
select * from @t
寻找将输出为
的转换下面是进行转换的当前脚本
select 'startinventory' as Activities,
bucket1=(select startinventory from @t where id=1),
bucket2=(select startinventory from @t where id=2),
bucket3=(select startinventory from @t where id=3),
bucket4=(select startinventory from @t where id=4) union all
select 'nocontact',
(select nocontact from @t where id=1),
(select nocontact from @t where id=2),
(select nocontact from @t where id=3),
(select nocontact from @t where id=4)
是否有更好的脚本编写方法?
答案 0 :(得分:1)
一个大的假设是您只有4个固定的存储桶,但如果这样做,它将做到这一点(提供示例数据)。
declare @bucket table(id int, bucket varchar(10), startinventory int, nocontact int)
insert @bucket values (1,'bucket1',1234,72500),(2,'bucket2',6784,60500),(3,'bucket3',678,52000),(4,'bucket4',234,45000)
select a as Activity,
sum(case when id=1 then case when a='startinventory' then startinventory else NoContact end else 0 end) as bucket1,
sum(case when id=2 then case when a='startinventory' then startinventory else NoContact end else 0 end) as bucket2,
sum(case when id=3 then case when a='startinventory' then startinventory else NoContact end else 0 end) as bucket3,
sum(case when id=4 then case when a='startinventory' then startinventory else NoContact end else 0 end) as bucket4
from (values ('startinventory'),('nocontact')) t(a)
cross join @bucket b
group by a