SQL Server：为非唯一值创建带有子节点的 XML答案

【问题标题】：SQL Server : creating XML with child nodes for non unique valuesSQL Server：为非唯一值创建带有子节点的 XML
【发布时间】：2014-07-22 22:26:28
【问题描述】：

我很难弄清楚如何从 SQL select 创建 XML 文件，该文件具有用于存储非唯一值的子节点。

我正在使用 Microsoft SQL Server 2012

例如

select company, founder
from companies

查询的结果是：

 microsoft corp       Bill Gates
 microsoft corp       Paul Allen
 apple inc            Steve Jobs
 apple inc            Ronald Wayne
 apple inc            Steve Wozniak

我想生成一个类似的 XML

<values>
       <company>microsoft corp</company>
       <founders>
           <founder id="1">Bill Gates</founder>
           <founder id="2">Paul Allen</founder>
       </founders>
</values>
<values>
       <company>apple inc</company>
       <founders>
           <founder id="1">Steve Jobs</founder>
           <founder id="2">Ronald Wayne</founder>
           <founder id="3">Steve Wozniak</founder>
        </founders>
       </company>
</values>

我不确定节点<founders> 是必需的，我认为让创始人直接在<values> 节点下也可以正常工作，因为他们留在正确的公司，获得一个ID列表。

我现在得到的是，使用FOR XML，然后我尝试了以下不同的选项：

<values>
       <company>apple inc</company>
       <founder>Steve Jobs</founder>
</values>
<values>
       <company>apple inc</company>
       <founder>Ronald Wayne</founder>
</values>
<values>
       <company>apple inc</company>
       <founder>Steve Wozniak</founder>
</values>

这与我需要实现的目标不兼容。

非常感谢任何帮助将所有创始人置于同一 <values> 节点下。

谢谢！

【问题讨论】：

刚刚更新很抱歉

标签： sql sql-server xml sql-server-2012

【解决方案1】：

上网搜索了一下，找到了解决办法：

DECLARE @table TABLE (company VARCHAR(100), founder VARCHAR(100))

 insert into @table SELECT 'microsoft corp', 'Bill Gates'
 insert into @table SELECT 'microsoft corp', 'Paul Allen'
 insert into @table SELECT 'apple inc', 'Steve Jobs'
 insert into @table SELECT 'apple inc', 'Ronald Wayne'
 insert into @table SELECT 'apple inc', 'Steve Wozniak'


 SELECT company, 
    (
        SELECT ROW_NUMBER()OVER( ORDER BY founder ) as 'founder/@id',
            founder as 'founder'
        FROM @table as F
        WHERE F.company = Comp.company
        FOR XML PATH(''), TYPE
    ) AS founders
 FROM @table as Comp
 GROUP BY company
 FOR XML PATH('VALUES')

结果：

<VALUES>
  <company>apple inc</company>
  <founders>
    <founder id="1">Ronald Wayne</founder>
    <founder id="2">Steve Jobs</founder>
    <founder id="3">Steve Wozniak</founder>
  </founders>
</VALUES>
<VALUES>
  <company>microsoft corp</company>
  <founders>
    <founder id="1">Bill Gates</founder>
    <founder id="2">Paul Allen</founder>
  </founders>
</VALUES>

【讨论】：

【解决方案2】：

测试数据

DECLARE @Companies TABLE (company VARCHAR(100), founder VARCHAR(100))
INSERT INTO @Companies VALUES
 ('microsoft corp',       'Bill Gates'),
 ('microsoft corp',       'Paul Allen'),
 ('apple inc',            'Steve Jobs'),
 ('apple inc',            'Ronald Wayne'),
 ('apple inc',            'Steve Wozniak')

查询

SELECT company
       ,(SELECT ROW_NUMBER() OVER (PARTITION BY Company ORDER BY Founder ASC)  AS [@ID] 
               ,B.founder [text()]
         FROM @companies B
        WHERE B.company = A.company
        FOR XML PATH('Founder'),TYPE) AS Founders
 FROM  @companies A
GROUP BY A.Company       
FOR XML PATH('values'),ROOT('Doc'), ELEMENTS

输出

<Doc>
  <values>
    <company>apple inc</company>
    <Founders>
      <Founder ID="1">Ronald Wayne</Founder>
      <Founder ID="2">Steve Jobs</Founder>
      <Founder ID="3">Steve Wozniak</Founder>
    </Founders>
  </values>
  <values>
    <company>microsoft corp</company>
    <Founders>
      <Founder ID="1">Bill Gates</Founder>
      <Founder ID="2">Paul Allen</Founder>
    </Founders>
  </values>
</Doc>

【讨论】：