Question

我有一个以行样式存储Profiles商店配置文件属性值的表，例如：

[ProfileID]     [PropertyDefinitionID]      [PropertyValue]
1               6                           Jone
1               7                           Smith
1               8                           Mr
1               3                           50000

和另一个属性定义表：

[PropertyDefinitionID]  [PropertyName]
6                       FirstName
7                       LastName
8                       Prefix
3                       Salary

如何使用PIVOT或任何其他方式以这种方式显示它：

[ProfileID] [FirstName] [LastName]  [Salary]
1           Jone        Smith       5000

Answer 1

没有PIVOT关键字，只需按分组

即可轻松完成此操作

select
    P.ProfileID,
    min(case when PD.PropertyName = 'FirstName' then P.PropertyValue else null end) as FirstName,
    min(case when PD.PropertyName = 'LastName' then P.PropertyValue else null end) as LastName,
    min(case when PD.PropertyName = 'Salary' then P.PropertyValue else null end) as Salary
from Profiles as P
    left outer join PropertyDefinitions as PD on PD.PropertyDefinitionID = P.PropertyDefinitionID
group by P.ProfileID

您也可以使用PIVOT关键字

执行此操作

select
    *
from
(
    select P.ProfileID, P.PropertyValue, PD.PropertyName
    from Profiles as P
        left outer join PropertyDefinitions as PD on PD.PropertyDefinitionID = P.PropertyDefinitionID
) as P
    pivot
    (
        min(P.PropertyValue)
        for P.PropertyName in ([FirstName], [LastName], [Salary])
    ) as PIV

更新：对于动态数量的属性 - 请查看Increment value in SQL SELECT statement

Answer 2

您可能需要将未知数量的PropertyName's转换为列。如果是这种情况，那么您可以使用动态sql生成结果：

DECLARE @cols AS NVARCHAR(MAX),
    @query  AS NVARCHAR(MAX)

select @cols = STUFF((SELECT distinct ',' + QUOTENAME(PropertyName) 
                    from propertydefinitions
            FOR XML PATH(''), TYPE
            ).value('.', 'NVARCHAR(MAX)') 
        ,1,1,'')

set @query = 'SELECT profileid, ' + @cols + ' from 
             (
                select p.profileid,
                  p.propertyvalue,
                  d.propertyname
                from profiles p
                left join propertydefinitions d
                  on p.PropertyDefinitionID = d.PropertyDefinitionID
            ) x
            pivot 
            (
                max(propertyvalue)
                for propertyname in (' + @cols + ')
            ) p '

execute(@query)

见SQL Fiddle with Demo。

T：SQL：从行中选择值作为列

2 个答案: