Question

我对SQL服务器“选择XML路径”查询很有经验，但现在我遇到了一个奇怪的问题。

以下查询正常工作：

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')

这将在此XML中产生（对于虚拟数据集）：

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

在我的结果（更大的一部分）声明中，我也需要第二个关键字：

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

所以我用union-select改变了我的（子）查询：

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')

但是现在我得到了错误“当EXISTS没有引入子查询时，只能在选择列表中指定一个表达式。”

我知道在子查询中有多个记录可能会导致多个元素的XML路径。但我不明白为什么不能用工会做到这一点。

有人能否指导我如何使用我的（子）查询中的2个关键字段完成XML？

非常感谢你。

Answer 1

您的子选择的问题是第一部分根本没有引用任何表（没有FROM-子句）。

此列表为我提供了您所要求的输出：

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

这是一个简化的示例，但这可以满足您的需求吗？

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')

SQL Server 2005在子选择问题中选择带有union的XML路径

2 个答案: