How to transform an XML using an XSLT?

Question

I have an XML file and an XSLT and I want to use the XSLT to trasform the XML into an HTML string which could be loaded into a TWebBrowser.

For my tests, I'm using these example files.

XML File:

<?xml version="1.0" encoding="UTF-8"?>
<breakfast_menu>

<food>
<name>Belgian Waffles</name>
<price>$5.95</price>
<description>Two of our famous Belgian Waffles with plenty of real maple syrup</description>
<calories>650</calories>
</food>

<food>
<name>Strawberry Belgian Waffles</name>
<price>$7.95</price>
<description>Light Belgian waffles covered with strawberries and whipped cream</description>
<calories>900</calories>
</food>

<food>
<name>Berry-Berry Belgian Waffles</name>
<price>$8.95</price>
<description>Light Belgian waffles covered with an assortment of fresh berries and whipped cream</description>
<calories>900</calories>
</food>

<food>
<name>French Toast</name>
<price>$4.50</price>
<description>Thick slices made from our homemade sourdough bread</description>
<calories>600</calories>
</food>

<food>
<name>Homestyle Breakfast</name>
<price>$6.95</price>
<description>Two eggs, bacon or sausage, toast, and our ever-popular hash browns</description>
<calories>950</calories>
</food>

</breakfast_menu>

XSLT File:

<?xml version="1.0" encoding="UTF-8"?>
<html xsl:version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<body style="font-family:Arial;font-size:12pt;background-color:#EEEEEE">
<xsl:for-each select="breakfast_menu/food">
  <div style="background-color:teal;color:white;padding:4px">
    <span style="font-weight:bold"><xsl:value-of select="name"/> - </span>
    <xsl:value-of select="price"/>
    </div>
  <div style="margin-left:20px;margin-bottom:1em;font-size:10pt">
    <p>
    <xsl:value-of select="description"/>
    <span style="font-style:italic"> (<xsl:value-of select="calories"/> calories per serving)</span>
    </p>
  </div>
</xsl:for-each>
</body>
</html>

Attempt 1:

I've found this solution and tried the function:

Uses
  XMLDoc, XMLIntf;

function Transform(XMLContent : string; XSLContent : string) : WideString;
var
  XML : IXMLDocument;
  XSL : IXMLDocument;
begin

  XML := LoadXMLData(XMLContent);
  XSL := LoadXMLData(XSLContent);

  XML.DocumentElement.TransformNode(XSL.DocumentElement, Result)

end;

But it produces an unexpected output:

Belgian Waffles$5.95Two of our famous Belgian Waffles with plenty of real maple syrup650Strawberry Belgian Waffles$7.95Light Belgian waffles covered with strawberries and whipped cream900Berry-Berry Belgian Waffles$8.95Light Belgian waffles covered with an assortment of fresh berries and whipped cream900French Toast$4.50Thick slices made from our homemade sourdough bread600Homestyle Breakfast$6.95Two eggs, bacon or sausage, toast, and our ever-popular hash browns950

Attempt 2:

I've found the following function in the "Using the MSXML Parser/ Transform Engine" section of this page (Note: it's a Delphi 5 example but I'm using Delphi 2007).

function DoTransform(const xml, xsl : string ): string;
var
  XMLDoc : IXMLDOMDocument;
  XSLDoc : IXMLDOMDocument;
  Template : IXSLTemplate;
  Processor : IXSLProcessor;
begin
  Result := '';
  try
    XMLDoc := CoFreeThreadedDOMDocument30.Create;
    XSLDoc := CoFreeThreadedDOMDocument30.Create;
    XMLDoc.load(xml);
    XSLDoc.load(xsl);
    Template := CoXSLTemplate30.Create;
    Template.stylesheet := XSLDoc;
    Processor := Template.createProcessor;
    Processor.input := XMLDoc;
    Processor.transform;
    result :=  Processor.output;
  finally
    XMLDoc := nil;
    XSLDoc := nil;
  end;
end;

I've imported the Microsoft XML type library...

Component -> Import Component -> Import a Type Library -> Next -> Microsoft XML, v6.0

...and added MSXML2_TLB to the uses clause but then some other errors occours:

E2003 Undeclared identifier: 'CoFreeThreadedDOMDocument30'

E2003 Undeclared identifier: 'CoXSLTemplate30'

I've switched CoFreeThreadedDOMDocument30 to CoFreeThreadedDOMDocument60 and CoXSLTemplate30 to CoXSLTemplate60 and now it compiles without errors.

At runtime, at this line:

Template.stylesheet := XSLDoc;

It raises the following exception (In italian):

Il foglio di stile non include un elemento documento. Il foglio di stile è vuoto oppure potrebbe essere un documento XML in formato non corretto.

In english it should be something like this:

The stylesheet doesn't include a document element. The stylesheet is empty or it could be a bad formatted XML document.

I've tested with other example files and the error is always the same and I don't understand which could be the problem.

Am I on the right way or are there better ways to do what I need?

Larsdk Larsdk · Accepted Answer · 2019-01-04T09:18:59

I have the following code that works. It is assuming a file containing the XSLT but that should be easy to change.

It was written a long time ago and I was in a hurry. So there might be room for some improvements. Hope it helps.

uses
  Windows, ComObj, XMLDoc, msxmldom, msxml;

function DOMToMSDom(const Doc: IDOMDocument): IXMLDOMDocument3;
begin
  Result := ((Doc as IXMLDOMNodeRef).GetXMLDOMNode as IXMLDOMDocument3);
end;

function TransformXMLDocWithXSLTFile(const Doc: XMLIntf.IXMLDocument; const StyleSheetLocation: string; var TransformedData, Error: string): Boolean;
var
  MsxmlDoc: IXMLDOMDocument3;
  xslStyle : IXMLDOMDocument;
begin
  Result := False;
  if not FileExists(StyleSheetLocation) then 
  begin
    Error := 'Specified XSLT stylesheet file does not exist: ' + StyleSheetLocation;
    Exit;
  end;
  try
    MsxmlDoc := DOMToMSDom(Doc.DOMDocument);
    xslStyle := CoDOMDocument60.Create;
    xslStyle.load(StyleSheetLocation);
    IXMLDOMDocument3(xslStyle).setProperty('AllowXsltScript', True);
    TransformedData := MsxmlDoc.transformNode(xslStyle);
    Result := True;
  except
    on E: Exception do 
    begin
      Error := E.Message;
    end;
  end;
end;

How to transform an XML using an XSLT?

3 Answers