1

基本上我需要以以下 XML 格式从 SQL Server 表中返回一些数据：

<querydata>
  <entity name="Person.Contact">
    <row>
      <field name="FirstName">Gustavo</field>
      <field name="LastName">Achong</field>
    </row>
    <row>
      <field name="FirstName">Catherine</field>
      <field name="LastName">Abel</field>
    </row>
...
  </entity>
</querydata>

我想出了以下 SQL 语句：

select 'Person.Contact' as "@name", 
(select FirstName, LastName from Person.Contact for XML path('row'), TYPE)
for XML path('entity'), root('querydata')

产生这个输出：

<querydata>
  <entity name="Person.Contact">
    <row>
      <FirstName>Gustavo</FirstName>
      <LastName>Achong</LastName>
    </row>
    <row>
      <FirstName>Catherine</FirstName>
      <LastName>Abel</LastName>
    </row>
....
  </entity>
</querydata>

但我没有更进一步。谢谢！

4

2 回答 2

1

您需要对数据进行反透视。

尝试使用子查询：

SELECT 'FirstName' as [@name], FirstName as [*]
union all
SELECT 'LastName' as [@name], LastName as [*]
for xml path('field')

或者类似的东西......

我没有 SQL（今天在我的 iPhone 上），但我正在考虑：

select 'Person.Contact' as "@name", 
(select (SELECT 'FirstName' as [@name], FirstName as [*]
union all
SELECT 'LastName' as [@name], LastName as [*]
for xml path('field')) from Person.Contact for XML path('row'), TYPE)
for XML path('entity'), root('querydata')
于 2010-01-23T02:11:25.167 回答
0

非常感谢罗布！你肯定让我走上了正确的道路，为你+1！我必须将所有内容都包装在 SELECT * FROM 语句中，否则 SQL 服务器会报错。这是最终的工作查询：

SELECT 'Person.Contact' as "@name",
(SELECT 
    (SELECT * from (SELECT 'FirstName' as [@name], [FirstName] as [*]
    union all
    SELECT 'LastName' as [@name], [LastName] as [*]) y
    for xml path('field'), TYPE)
from Person.Contact for XML path, TYPE)
for XML path('entity'), root('querydata')
于 2010-01-23T02:58:06.187 回答