添加额外的节点 XML答案

【问题标题】：Add extra node XML添加额外的节点 XML
【发布时间】：2017-03-30 13:58:39
【问题描述】：

我正在使用 FOR XML PATH 提取一些数据。但是，我无法获得节点。预期结果：

<Users>
  <User UserId="1">
   <Profile>
     <PersonalDetails>
        <FirstName>Name1</FirstName>
        <LastName>Last1</LastName>
     </PersonalDetails>
     <Reward>0</Reward>
   </Profile>
 </User>
 <User UserId="2">
    <Profile>
      <PersonalDetails>
          <FirstName>Name2</FirstName>
          <LastName>Last2</LastName>
      </PersonalDetails>
      <Reward>0</Reward>
    </Profile>
 </User>
 <User UserId="3">
   <Profile>
      <PersonalDetails>
        <FirstName>Name3</FirstName>
        <LastName>Last3</LastName>
      </PersonalDetails>
      <Reward>0</Reward>
   </Profile>
 </User>
 <User UserId="4">
   <Profile>
    <PersonalDetails>
        <FirstName>Name4</FirstName>
        <LastName>Last4</LastName>
    </PersonalDetails>
    <Reward>0</Reward>
   </Profile>
 </User>
</Users>

让我用这个例子告诉你：

DECLARE @User TABLE ( ID INT
                 ,FirstName VARCHAR(20)
                 ,LastName VARCHAR(20)
                 ,Reward BIT
                 ) 

INSERT INTO @User (ID, FirstName, LastName, Reward )
VALUES (1, 'Name1', 'Last1', 0)
      ,(2, 'Name2', 'Last2', 0)
      ,(3, 'Name3', 'Last3', 0)
      ,(4, 'Name4', 'Last4', 0)


    SELECT ID AS '@UserId'
            ,( SELECT FirstName AS 'FirstName'
                     ,LastName AS 'LastName'
               FROM @User
               WHERE ID = users.ID
               FOR XML PATH('PersonalDetails'), TYPE)       
           ,Reward AS 'Reward'
    FROM @User AS users
    FOR XML PATH('User'), ROOT('Users'), ELEMENTS

这是我目前所得到的：

 <Users>
<User UserId="1">
   <PersonalDetails>
      <FirstName>Name1</FirstName>
      <LastName>Last1</LastName>
   </PersonalDetails>
   <Reward>0</Reward>
</User>
</User>
   <User UserId="2">
     <PersonalDetails>
       <FirstName>Name2</FirstName>
       <LastName>Last2</LastName>
     </PersonalDetails>
     <Reward>0</Reward>
   </User>
   <User UserId="3">
      <PersonalDetails>
         <FirstName>Name3</FirstName>
         <LastName>Last3</LastName>
      </PersonalDetails>
      <Reward>0</Reward>
   </User>
   <User UserId="4">
     <PersonalDetails>
       <FirstName>Name4</FirstName>
       <LastName>Last4</LastName>
     </PersonalDetails>
     <Reward>0</Reward>
   </User>
 </Users>

我无法添加额外的 Profile 元素。我试过使用 modify('insert as first into (/Users/User)[1]') 但输出也没有那么快。你能帮我解决这个问题吗？

【问题讨论】：

这是一个非常好的问题！独立测试场景，明确的问题，自己的尝试，错误和预期的输出。太好了，投了赞成票！
感谢@Shnugo :)

标签： sql xml sql-server-2016 for-xml-path

【解决方案1】：

您的数据在内部似乎是普通的1:1：嵌套，但不是1:n。这不需要子选择。使用FOR XML PATH，您可以直接指定嵌套：

DECLARE @User TABLE ( ID INT
                 ,FirstName VARCHAR(20)
                 ,LastName VARCHAR(20)
                 ,Reward BIT
                 ) 

INSERT INTO @User (ID, FirstName, LastName, Reward )
VALUES (1, 'Name1', 'Last1', 0)
      ,(2, 'Name2', 'Last2', 0);

SELECT ID AS [@UserId]
      ,FirstName AS [Profile/PersonalDetails/FirstName]  
      ,LastName AS [Profile/PersonalDetails/LastName]  
      ,Reward AS [Profile/Reward]
FROM @User AS u
FOR XML PATH('User'),ROOT('Users')

结果

<Users>
  <User UserId="1">
    <Profile>
      <PersonalDetails>
        <FirstName>Name1</FirstName>
        <LastName>Last1</LastName>
      </PersonalDetails>
      <Reward>0</Reward>
    </Profile>
  </User>
  <User UserId="2">
    <Profile>
      <PersonalDetails>
        <FirstName>Name2</FirstName>
        <LastName>Last2</LastName>
      </PersonalDetails>
      <Reward>0</Reward>
    </Profile>
  </User>
</Users>

【讨论】：