我的查询是:
SELECT vendor.id, insurances.id AS ins_id, vendor_insurances.expiry_date
FROM vendor
INNER JOIN vendor_insurances
ON vendor.id=vendor_insurances.vendor_id
和输出:
id ins_id expiry_date
================================
28 1 2006-01-01
28 11 2008-01-01
我想将其转换为:
id 1 11
======================================
28 2006-01-01 2008-01-01
谢谢,
您将需要使用PIVOT和做同样的事情到这一点:
静态透视 - 为将要转动列短短数
select *
from
(
SELECT v.id
, vi.id AS ins_id
, vi.expiry_date
FROM vendor v
INNER JOIN vendor_insurances vi
ON v.id=vi.vendorId
) x
PIVOT
(
MIN(expiry_date)
FOR ins_id IN ([1], [11])
) p
请参阅SQL小提琴的工作演示
或者你可以使用动态透视,如果你有很多的项目PIVOT的:
DECLARE @cols AS NVARCHAR(MAX),
@query AS NVARCHAR(MAX);
select @cols = STUFF((SELECT distinct ',' + QUOTENAME(vi.id)
FROM vendor v
INNER JOIN vendor_insurances vi
ON v.id=vi.vendorId
FOR XML PATH(''), TYPE
).value('.', 'NVARCHAR(MAX)')
,1,1,'')
set @query = 'SELECT id, ' + @cols + ' from
(
SELECT v.id
, vi.id AS ins_id
, vi.expiry_date
FROM vendor v
INNER JOIN vendor_insurances vi
ON v.id=vi.vendorId
) x
pivot
(
MIN(expiry_date)
for ins_id in (' + @cols + ')
) p '
execute(@query)
这两种会给你同样的结果。