Hi there,
I have a problem with transforming from XML to CSV using a very simple stylesheet.

Following is an example input file
------------------------------------
<CD>
 <PROJECT ID="TEST" CODE="TESTCODE"></PROJECT>
 <PROJECT ID="TEST2" CODE="TESTCODE2"></PROJECT>
</CD>

Following is the VB 6 code
----------------------------

Public Function TransformFromDOM(xmlSource As MSXML2.FreeThreadedDOMDocument40, strXSL As String) As FreeThreadedDOMDocument40
    On Error GoTo ErrHandler

    Dim xmlOutput As New MSXML2.FreeThreadedDOMDocument40
    Dim xmlStylesheet As New MSXML2.FreeThreadedDOMDocument40
    Dim xslTemplate As New MSXML2.XSLTemplate40
    Dim xslProcessor As MSXML2.IXSLProcessor

    xmlStylesheet.Load strXSL

    If xmlStylesheet.parseError.reason <> "" Then
        MsgBox "ERROR: " & xmlStylesheet.parseError.reason & vbCrLf & "SOURCE: " & _
                    xmlStylesheet.parseError.srcText, vbExclamation, App.Title
        Exit Function
    End If

    Set xslTemplate.stylesheet = xmlStylesheet

    Set xslProcessor = xslTemplate.createProcessor

    xslProcessor.input = xmlSource
    xslProcessor.output = xmlOutput

    xslProcessor.Transform

    Set TransformFromDOM = xmlOutput

    Exit Function

ErrHandler:
    MsgBox "ERROR: " & err.Number & " " & err.Description, _
                vbCritical, App.Title
    Set TransformFromDOM = Nothing
End Function


Now here is the simple transformation which isn't working at all
-----------------------------------------------------------------
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output version="1.0" indent="no" method="text" omit-xml-declaration="yes" />

      <xsl:template match="/">
              <xsl:apply-templates select="CD"/>
    </xsl:template>

    <xsl:template match="CD"><xsl:text>Ref,Start,Finish,Title#13;</xsl:text>
        <xsl:for-each select="//Project">
            <xsl:value-of select="concat( @ID, ',', @CODE)"/>
            <xsl:text>#13;</xsl:text>
        </xsl:for-each>
    </xsl:template>
</xsl:stylesheet>


The output being generated is a blank file completely.

Does anyone know what is happening.. Note that when I apply the transformation from within XML Spy using the same XML parser (MSXML4.0) everything works fine.

Thankx in advance
Jonathan

A few things about your stylesheet:
1. XML is case-sensistive so you need to select/match PROJECT not Project
2. there is no need to use the expensive // operation to match the PROJECT node
3. a new-line is written like this #13;#10;
4. there is no need for the concat() function - just output the items and output a comma between them
try this stylesheet and see what happens
I'm surprised it works in XML Spy!

hth
Phil

How do we get an ampersand to show up in this forum?

The new-line should be [ampersand]#13;[ampersand]#10;

rgds
Phil

Just noticed a couple of things about your VB
1. "xmlStylesheet.Load strXSL" should read "xmlStylesheet.LoadXML strXSL" because the Load methos is for loading files
2. you are setting the output of your transformation to a DOM Document, but this will not work because the output is text not XML. You need to output to a string. The way to do that is do not set the output property, but instead read it into a string variable after the transform has been done. Like this:
hth
Phil