NicoAgenci wrote:


> private void Page_Load(object sender, System.EventArgs e)
> 
> {
> 
> String xmlPath = Server.MapPath("/data/page.xml") ;
> 
> String xslPath = Server.MapPath("/xsl/i_infoPage.xsl") ;
> 
> String strUrlPage = "/default.aspx";
> 
> String xpathPage = "page[url='" + strUrlPage + "']";
> 
> 
> // Nouvo XmlDoc pour l'affichage des Infos de Page
> 
> XmlDocument xmldocInfo = new XmlDocument();
> 
> xmldocInfo.Load(xmlPath);
> 
> XslTransform xsltInfo = new XslTransform();
> 
> xsltInfo.Load(xslPath);
> 
> // Crée un Arg pour le passage en Xslt du Composant Asp:Xml
> 
> XsltArgumentList argsInfo = new XsltArgumentList();
> 
> argsInfo.AddParam("urlPage","", "'Nico'");
> 
> // Réalise la transformation
> 
> xsltInfo.Transform(xmldocInfo, argsInfo);

Well, above line is superfluous. Technically here you are starting XSL 
transformation and not using its result. asp:xml control will do 
transformation for you, that's its the only function. You only need to 
provide source document, xsl stylesheet and optional parameters.

> // Source Xml et Xslt pour le Composant AspXml
> 
> aspxmlInfo.Document = xmldocInfo;
> 
> aspxmlInfo.Transform = xsltInfo;

Here is how you can provide parameters to asp:xml control:

aspxmlInfo.TransformArgumentList = argsInfo;

-- 
Oleg Tkachenko [XML MVP]
http://blog.tkachenko.com