Transform - Sorting xml document using vb .net

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  11 years ago

  by gixxer05

  Jermaine Major

  Washington, United States

  Joined 11 years ago

  I need help creating a function for transforming an xml document using xslt. I am able to copy the document using xsl:copy-of, however it's not copying the root of the source xml document. As shown below in italic text, the root of the xml should be :

       <?xml version="1.0" encoding="UTF-8" ?> 
  - <!-- Sample XML file generated by XMLSpy v2008 sp1 (http://www.altova.com)
    -->
  

  * - * - -

  The xslt transformation function I have follows:

        Dim xmldocumentobject As XmlDocument
              xmldocumentobject = New XmlDocument
              Dim strfilepath As String = xmlFilePath
  
              'Load the unsorted/current Xml file
              xmldocumentobject.Load(strfilepath)
  
          'Insert function to overwrite previous unsorted xmldocument with newly sorted newxmldocument
            Dim str_sortedxmldoc As String                           
            str_sortedxmldoc = XsltSortDoc(xmldocumentobject.ToString)
  
             ' Re-Load the transformed/sorted xml document into the xml document object
            xmldocumentobject.LoadXml(str_sortedxmldoc)
            'Then save xml document in configured file path
            xmldocumentobject.Save(xmlFilePath)
  
          Public Function XsltSortDoc(ByVal sXMLDocument As String) As String
          Dim xd As XmlDocument
          Dim xdNav As XPathNavigator
          Dim tr As Xsl.XslCompiledTransform
          Dim sw As StringWriter
  
  
          xd = New XmlDocument()
          xd.LoadXml(sXMLDocument)
          xdNav = xd.CreateNavigator()
          tr = New Xsl.XslCompiledTransform()
          tr.Load(config.xsltFilePath)
          sw = New StringWriter()
          tr.Transform(xdNav, Nothing, sw)
  
          Return sw.ToString()
  
      End Function
  

  The xslt file defining the transformation follows:

    <?xml version="1.0" encoding="utf-8"?>
  <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
      xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl">
    <xsl:output method="xml" indent="yes"/>
  
    <!--Copy List and all nodes, namespaces, child nodes, and text from source xml doc-->
    <!-- by using @*|*|text()|comment()|processing-instruction() or @* | node()-->
    <xsl:template match="/">
      <xsl:copy-of select="@* | node()"/>
    </xsl:template>
  
  <!--Match the Slug nodes 
    Copy the current node's top-level values(the tag and it's attributes,
           but not it's descendents)-->
    <xsl:template match="List">
      <xsl:copy>
         <!--Apply transformation to all S nodes.-->
        <xsl:apply-templates select="S">
          <!--For all Slugname attributes, sort them 
                ascendingly according to their @SlugName attribute.-->
          <xsl:sort select="@SName" order="ascending"/>
          <!--Copy the entire node (include its descendantas)-->
         </xsl:apply-templates>
      </xsl:copy>
  </xsl:template>
  </xsl:stylesheet>
  

  The above code produces the following outputed xml file in the debugger. Notice that the has been commented out by the transformation. This causes the transformation function to fail during runtime compilation. :

    <?xml version="1.0" encoding="utf-8"?>
  <!--Sample XML file generated by XMLSpy v2008 sp1 (http://www.altova.com)-->
  

  * *

  <B BID="SU">
      <O OID="MN">
    <S SName="BFirstTEST" Description="Test PDF" /> 
    <S SName="ASecondTEST" Description="Test PDF" /> 
      </O>
    </B>
  

  My IDE is Visual Basic .Net 2008. I need to reproduce the xml file with the SlugName attributes sorted in ascending order. Any help is appreciated. Thanks.

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• • Reply

  11 years ago

  by gixxer05

  Jermaine Major

  Washington, United States

  Joined 11 years ago

  The root node which is being commented out is as follows:

  <List xsi:noNamespaceSchemaLocation="SlugList.xsd" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
  

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