sql – 带xmlns的OPENXML：dt

使用OPENXML在MSSQL 2005中获取dt元素.
如何在xml中获取xmlns：dt元素？例如,获取列出产品ID和国家/地区代码的两行结果集.

121403 GBR

121403美国

declare @xmldata xml
    set @xmldata = 
    '<?xml version="1.0"?>
    <data xmlns="http://www.aaa.com/master_browse_response" xmlns:dt="http://www.aaa.com/DataTypes">
      <products>
        <product>
       <product_id><![CDATA[121403]]></product_id>
          <countries>
            <dt:country>GBR</dt:country>
            <dt:country>USA</dt:country>
          </countries>
        </product>
     </products>
    </data>'

     DECLARE @hDoc int,@rootxmlns varchar(100)
     SET @rootxmlns = '<root xmlns:hm="http://www.aaa.com/master_browse_response"/>'

     EXEC sp_xml_preparedocument @hDoc OUTPUT,@xmldata,@rootxmlns  

     SELECT *
     FROM OPENXML(@hDoc,'//hm:product',2)
     WITH ([hm:product_id] int,[hm:countries] varchar(100))

     --clean up 
     EXEC sp_xml_removedocument @hDoc

这是我通过使用xmlEdgeTable知道的一个解决方案,但我正在寻找更好的解决方案.

DECLARE @hDoc int,@rootxmlns varchar(100)
SET @rootxmlns = '<root xmlns:hm="http://www.aaa.com/master_browse_response"/>'

EXEC sp_xml_preparedocument @hDoc OUTPUT,@rootxmlns  

CREATE TABLE #xmlEdgeTable
( 
    id int,parentid int,localname varchar(20),[text] varchar(20)
)

INSERT INTO #xmlEdgeTable
SELECT id,parentid,localname,cast([text] as varchar(20)) 
FROM OPENXML(@hDoc,2)

SELECT t6.text,t2.text FROM #xmlEdgeTable AS t1 INNER JOIN 
    #xmlEdgeTable AS t2 ON t1.id = t2.parentid INNER JOIN 
    #xmlEdgeTable AS t3 ON t3.id = t1.parentid INNER JOIN 
    #xmlEdgeTable AS t4 ON t4.id = t3.parentid INNER JOIN 
    #xmlEdgeTable AS t5 ON t4.id = t5.parentid INNER JOIN
    #xmlEdgeTable AS t6 ON t5.id = t6.parentid 
WHERE t1.localname = 'country' and t5.localname ='product_id'

--clean up 
EXEC sp_xml_removedocument @hDoc
DROP TABLE #xmlEdgeTable

解决方法

declare @xmldata xml    
set @xmldata = 
'<data xmlns="http://www.aaa.com/master_browse_response" xmlns:dt="http://www.aaa.com/DataTypes">
  <products>
    <product>
      <product_id>121403</product_id>
      <countries>
        <dt:country>GBR</dt:country>
        <dt:country>USA</dt:country>
      </countries>
    </product>
  </products>
</data>'

;WITH XMLNAMESPACES 
(
    DEFAULT 'http://www.aaa.com/master_browse_response','http://www.aaa.com/DataTypes' as dt
)
SELECT  x.c.value('(../../product_id)[1]','varchar(100)') as product_id,x.c.value('(.)[1]','varchar(100)') as country
FROM @xmldata.nodes('/data/products/product/countries/dt:country') x(c)

较新的XQUERY功能是解决问题的更好选择.

编辑：
与OPENXML相同的解决方案是：

declare @xmldata xml    
set @xmldata = 
'<data xmlns="http://www.aaa.com/master_browse_response" xmlns:dt="http://www.aaa.com/DataTypes">
  <products>
    <product>
      <product_id>121403</product_id>
      <countries>
        <dt:country>GBR</dt:country>
        <dt:country>USA</dt:country>
      </countries>
    </product>
  </products>
</data>'

DECLARE @hDoc int,@rootxmlns varchar(100)
SET @rootxmlns = '<root xmlns:hm="http://www.aaa.com/master_browse_response" xmlns:dt="http://www.aaa.com/DataTypes"/>'
EXEC sp_xml_preparedocument @hDoc OUTPUT,@rootxmlns

SELECT *
FROM OPENXML(@hDoc,'//hm:product/hm:countries/dt:country',2)
        WITH(Country    varchar(100) '.',Product_ID varchar(100) '../../hm:product_id')

EXEC sp_xml_removedocument @hDoc

（编辑：PHP编程网 - 黄冈站长网）

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