Excel VBA Coding for xls transformation along with Paramters

Question

I need to parse xsl along with its parameters using VBA code. I can use VBA code from the below link(for your reference) but only thing is I need to pass XSLT Parameters through VBA code.

VBA code : Hperlink

My (ds_test.xsl) file

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:output method="xml" indent="yes" />
<xsl:param name="job" />
<xsl:param name="src" />
<xsl:param name="spath" />
<xsl:template match="/">
  <DSExport>
     <Job>
        <xsl:attribute name="Identifier">
           <xsl:value-of select="$job" />
        </xsl:attribute>
        <Record Identifier="V25S0P1" Type="CustomOutput" Readonly="0">
           <Collection Name="Columns" Type="OutputColumn">
              <xsl:copy-of select="document($src)//Record[@Identifier=$spath]//SubRecord" />
           </Collection>
        </Record>
     </Job>
  </DSExport>
</xsl:template>
</xsl:stylesheet>

My input (Metadata.xml) file

<?xml version="1.0" encoding="UTF-8"?>
<DSExport>
<Header CharacterSet="CP1252" ExportingTool="IBM InfoSphere DataStage Export" ToolVersion="8" ServerName="HCL-BOEING-DS" ToolInstanceID="EFBI_BAL_OPT" Date="2014-01-21" Time="19.09.04" ServerVersion="9.1" />
<TableDefinitions>
  <Record Identifier="TEST1" DateModified="2013-12-23" TimeModified="11.01.03" Type="MetaTable" Readonly="0">
     <Collection Name="Columns" Type="MetaColumn">
        <SubRecord>
           <Property Name="Name">BEMSID</Property>
           <Property Name="Description">BEMSID: string[max=10]</Property>
           <Property Name="SqlType">12</Property>
           <Property Name="Precision">10</Property>
        </SubRecord>
        <SubRecord>
           <Property Name="Name">EMPL_NM</Property>
           <Property Name="Description">EMPL_NM: string[max=18]</Property>
           <Property Name="SqlType">12</Property>
           <Property Name="Precision">18</Property>
        </SubRecord>
     </Collection>
  </Record>
</TableDefinitions>
</DSExport>

My Expected Output in format of XML (output.xml)

Note : Here is my question - In runtime I should pass XSL parameter's value through Excel VBA code as I mentioned earlier. Let us assume that I am giving the below parameter value

VBA code : Hperlink

$job = "PXJ_TEST1"
$src = "Metadata.xml"
$spath = "TEST1"

output.xml should be in below format

<?xml version="1.0" encoding="UTF-8"?>
<DSExport>
<Job Identifier="PXJ_TEST1">
    <Record Identifier="V25S0P1" Type="CustomOutput" Readonly="0">
        <Collection Name="Columns" Type="OutputColumn">
            <SubRecord>
                <Property Name="Name">BEMSID</Property>
                <Property Name="Description">BEMSID: string[max=10]</Property>
                <Property Name="SqlType">12</Property>
                <Property Name="Precision">10</Property>
            </SubRecord>
            <SubRecord>
                <Property Name="Name">EMPL_NM</Property>
                <Property Name="Description">EMPL_NM: string[max=18]</Property>
                <Property Name="SqlType">12</Property>
                <Property Name="Precision">18</Property>
            </SubRecord>
        </Collection>
    </Record>
</Job>
</DSExport>

Parfait · Accepted Answer · 2019-06-07 15:56:11Z

For future readers who land on this page, below is a concrete implementation of passing parameters to XSLT from VBA using OP's inputs:

Sub XSLTransform()
    ' REFERENCE Microsoct XML, v6.0
    Dim xmlDoc As New MSXML2.DOMDocument60, newDoc As New MSXML2.DOMDocument60
    Dim xslDoc As New MSXML2.FreeThreadedDOMDocument60
    Dim xslTemp As New MSXML2.XSLTemplate60
    Dim xslProc As Object

    ' LOAD XML AND XSL FILES
    xmlDoc.async = False
    xmlDoc.Load "C:\Path\To\Input.xml"

    xslDoc.async = False
    xslDoc.setProperty "AllowDocumentFunction", True
    xslDoc.Load "C:\Path\To\XSLT_Script.xsl"

    ' INITIALIZE NEEDED OBJECTS
    Set xslTemp.stylesheet = xslDoc
    Set xslProc = xslTemp.createProcessor()

    With xslProc
       .input = xmlDoc
       .addParameter "job", "PXJ_TEST1"             ' ADD PARAMETER(S)
       .addParameter "src", "MultiParamXSLT.xml"
       .addParameter "spath", "TEST1"

       .transform                                   ' TRANSFORM XML
       newDoc.LoadXML .output                       ' LOAD RESULT TREE
       newDoc.Save "C:\Path\To\Output.xml"          ' SAVE OUTPUT TO FILE
    End With

    Set xmlDoc = Nothing: Set newDoc = Nothing
    Set xslDoc = Nothing: Set xslTemp = Nothing: Set xslProc = Nothing
End Sub

Martin Honnen · Accepted Answer · 2014-01-29 13:14:07Z

0

If you need to set parameters then the simple approach with transformNode does not work, you need to use the API presented in http://msdn.microsoft.com/en-us/library/ms762799%28v=vs.85%29.aspx (the page only has JScript and C++ unfortunately but of course you can use the same API and objects with VBScript or VBA). Then you can use the method http://msdn.microsoft.com/en-us/library/ms762312%28v=vs.85%29.aspx addParameter to set a parameter.

answered Jan 29, 2014 at 13:14

Martin Honnen

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