SQL:动态视图,其列名基于源表中的列值

时间:2012-06-11 19:05:55

标签: sql sql-server sql-server-2005

在这里给出两个样本表:

门票表

ID  User    Description

0   James   This is a support ticket
1   Fred    This is a ticket too

属性表

ID  TicketID    Label           Value

0   0           Engineer        Scott
1   1           Engineer        Dale
2   0           Manu            Dell
3   1           Manu            HP
4   0           OS              Windows
5   1           OS              Linux

我怎样才能达到这样的观点:

ID  User    Description                 Engineer    Manu    OS

1   James   This is a support ticket    Scott       Dell    Windows
2   Fred    This is a ticket too        Dale        HP      Linux

请务必注意,属性表并不总是相同。一些“门票”可能具有其他人没有的属性。

这甚至可能吗?

1 个答案:

答案 0 :(得分:14)

您可以使用PIVOT执行此操作。在执行PIVOT时,您可以使用以下两种方法之一:使用Static Pivot编写要转换的行,或使用Dynamic Pivot在运行时创建列列表:

Static Pivot(参见SQL Fiddle for Demo):

select id, [user], [engineer], [manu], [OS]
from 
(
    select t.id
        , t.[user]
        , p.ticketid
        , p.label
        , p.value
    from tickets t
    inner join properties p
        on t.id = p.ticketid
) x
pivot
(
    min(value)
    for label in ([engineer], [manu], [OS])
) p

或者您可以使用动态数据透视(请参阅SQL Fiddle for Demo):

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

select @cols = STUFF((SELECT distinct ',' + QUOTENAME(p.label) 
                    from tickets t
                    inner join properties p
                        on t.id = p.ticketid
            FOR XML PATH(''), TYPE
            ).value('.', 'NVARCHAR(MAX)') 
        ,1,1,'')

set @query = 'SELECT id, [user], ' + @cols + ' from 
             (
                 select t.id
                        , t.[user]
                        , p.ticketid
                        , p.label
                        , p.value
                    from tickets t
                    inner join properties p
                        on t.id = p.ticketid
            ) x
            pivot 
            (
                min(value)
                for label in (' + @cols + ')
            ) p '

execute(@query)

两个查询都会返回相同的结果。