SQL Server 2005中的XML路径与子选择问题联合选择

Question

我而经历了SQL服务器“选择XML路径”查询，但现在我遇到了一个奇怪的问题。

下面的查询工作正常：

select 
(
     select
     'Keyfield1' as "@Name",
    t1.Keyfield1 as "Value"
    from MyTable t1
    where 
    t1.KeyField1= t2.KeyField1 and
    t1.KeyField2= t2.KeyField2
    for xml path('Field'),type, elements 
) as 'Key'
from MyTable t2
for XML path('Path') , elements XSINIL, root('Root')

这将导致（对于虚设数据集）在这个XML：

<Root>  
  <Path>
    <Key Name="KeyField1">
      <Value>DummyValue1</Value>
    </Key>
  </Path>
</Root>

在我发言我需要的第二关键字段太这（一个更大的部分）的结果：

<Root>  
  <Path>
    <Key Name="KeyField1">
      <Value>DummyValue1</Value>
    </Key>
    <Key Name="KeyField2">
      <Value>DummyValue2</Value>
    </Key>
  </Path>
</Root>

，所以我改变我的（亚）查询与工会选择为：

select 
(
     select
     'Keyfield1' as "@Name",
    t1.Keyfield1 as "Value"
     union all
     select
     'Keyfield2' as "@Name",
    t1.Keyfield2 as "Value"
    from MyTable t1
    where 
    t1.KeyField1= t2.KeyField1 and
    t1.KeyField2= t2.KeyField2
    for xml path('Field'),type, elements 
) as 'Key'
from MyTable t2
for XML path('Path') , elements XSINIL, root('Root')

但现在我得到错误的“只有一个表达式可以在选择列表中指定当子查询不与EXISTS引入的。” 的

我知道它有可能具有与用于在多个元素的XML路径女巫结果的子查询的多个记录。但我不明白为什么这不能与工会进行。

有人可以把我在正确的方向如何accomplisch与2个KeyFields中的XML在我的（子）查询？

感谢名单大家。

Answer 1

与子选择的问题是，所述第一部分完全没有（无FROM-clause）参照的任何表。

此房源给我你的要求的输出：

declare @mytable table (
keyfield1 nvarchar(20),
keyfield2 nvarchar(20)
)

insert into @mytable values ('Dummyvalue1', 'Dummyvalue2')
select * from @mytable

select 
(
     select
     'Keyfield1' as "@Name",
    t1.Keyfield1 as "Value"
    from @mytable t1
    where 
    t1.KeyField1= t2.KeyField1 and
    t1.KeyField2= t2.KeyField2
    for xml path('Field'),type, elements 
) as 'Key'
from @mytable t2
for XML path('Path') , elements XSINIL, root('Root')


select 
(
    select * from (
      select
     'Keyfield1' as "@Name",
    t1.Keyfield1 as "Value"
    from @MyTable t1
    where 
    t1.KeyField1= t2.KeyField1
     union all
     select
     'Keyfield2' as "@Name",
    t3.Keyfield2 as "Value"
    from @MyTable t3
    where 
    t3.KeyField2= t2.KeyField2) a
    for xml path('Field'),type, elements 
) as 'Key'
from @MyTable t2
for XML path('Path') , elements XSINIL, root('Root')

Answer 2

下面是一个简单的例子，但这并得到你所需要的？

select 
    (
        select
            'Keyfield1' as "@Name",
            'Blah' as "Value"
        for xml path('Key'),type, elements 
    ),
    (
        select
            'Keyfield2' as "@Name",
            'Blah' as "Value"
        for xml path('Key'),type, elements 
    )
for XML path('Path') , elements XSINIL, root('Root')