OPENXML with xmlns:dt | 易学教程

Use OPENXML to get dt element in MSSQL 2005. How can I get xmlns:dt element in xml? For example, get a result set of two rows that list product id and country code.

121403 GBR

121403 USA

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

Here is one solution that I know by using xmlEdgeTable, but I am looking for a better solution.

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  

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, '//hm:product',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

Jonathan Kehayias

Is there a particular reason that you need to use OPENXML to do this? You can easily get the information with a XQUERY in 2005 like this:

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)

The newer XQUERY capabilities are a much better choice for solving your problem.

EDIT: The same solution with OPENXML would be:

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, @xmldata, @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

Vincent

For a small dataset, there is no big difference between XQuery and OPENXML

Here are results to parse a 6.5 MB xml file to get 27,615 rows: