如何从具有动态节点数的XML文件创建SQL表?

Ser*_*gey 8 xml sql-server openxml

我正在使用SQL Server 2008.
任务:获取XML文件并将其解析为(n)SQL表.
问题:列数及其名称将根据XML而有所不同.

这是一些代码:

DECLARE @xmlData XML;  
SET @xmlData = '<root>
  <item id="1">
    <item_number>IT23</item_number>
    <title>Item number twenty-three</title>
    <setting>5 to 20</setting>
    <parameter>10 to 16</parameter>
  </item>
  <item id="2">
    <item_number>RJ12</item_number>
     <title>Another item with a 12</title>
     <setting>7 to 35</setting>
     <parameter>1 to 34</parameter>
  </item>
  <item id="3">
     <item_number>LN90</item_number>
     <title>LN with 90</title>
     <setting>3 to 35</setting>
     <parameter>9 to 50</parameter>
  </item>
</root>'
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例如,使用上面的XML,我需要返回一个如下所示的SQL表:

表快照

以下是我如何得到上表:

   DECLARE @idoc INT;
   EXEC sp_xml_preparedocument @idoc OUTPUT, @xmlData

   SELECT * 
   FROM  OPENXML (@idoc, '/root/item', 2)
   WITH (item_number VARCHAR(100),
         title VARCHAR(100),
         setting VARCHAR(100),
         parameter VARCHAR(100))
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现在让我们说XML改变了每个项节点都有一个名为'new_node'的新子节点.像这样:

<root>
  <item id="1">
    <item_number>IT23</item_number>
    <title>Item number twenty-three</title>
    <setting>5 to 20</setting>
    <parameter>10 to 16</parameter>
    <new_node>data</new_node>
  </item>
  <item id="2">
    <item_number>RJ12</item_number>
    <title>Another item with a 12</title>
    <setting>7 to 35</setting>
    <parameter>1 to 34</parameter>
    <new_node>goes</new_node>
  </item>
  <item id="3">
    <item_number>LN90</item_number>
    <title>LN with 90</title>
    <setting>3 to 35</setting>
    <parameter>9 to 50</parameter>
    <new_node>here</new_node>
  </item>
</root>
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我必须更改我的代码以包含新节点:

   SELECT * 
   FROM  OPENXML (@idoc, '/root/item', 2)
   WITH (item_number VARCHAR(100),
         title VARCHAR(100),
         setting VARCHAR(100),
         parameter VARCHAR(100),
         new_node VARCHAR(100))
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要获得此表:

XML的第二个表

所以问题是'item'的子节点会有所不同.

如何在不指定列的情况下生成相同的表?除了必须使用OPENXML之外还有其他方法吗?

Mik*_*son 12

使用动态数量的列,您需要动态SQL.

declare @XML xml = 
'<root>
  <item id="1">
    <item_number>IT23</item_number>
    <title>Item number twenty-three</title>
    <setting>5 to 20</setting>
    <parameter>10 to 16</parameter>
    <new_node>data</new_node>
  </item>
  <item id="2">
    <item_number>RJ12</item_number>
    <title>Another item with a 12</title>
    <setting>7 to 35</setting>
    <parameter>1 to 34</parameter>
    <new_node>goes</new_node>
  </item>
  <item id="3">
    <item_number>LN90</item_number>
    <title>LN with 90</title>
    <setting>3 to 35</setting>
    <parameter>9 to 50</parameter>
    <new_node>here</new_node>
  </item>
</root>'

declare @SQL nvarchar(max) = ''
declare @Col nvarchar(max) = ', T.N.value(''[COLNAME][1]'', ''varchar(100)'') as [COLNAME]' 

select @SQL = @SQL + replace(@Col, '[COLNAME]', T.N.value('local-name(.)', 'sysname'))
from @XML.nodes('/root/item[1]/*') as T(N)

set @SQL = 'select '+stuff(@SQL, 1, 2, '')+' from @XML.nodes(''/root/item'') as T(N)' 

exec sp_executesql @SQL, N'@XML xml', @XML
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  • 我认为这是我见过的最好的TSQL/XML技巧之一,它解决了我的问题 - bravo (2认同)