Subject: Re: [xsl] document-uri in MSXML From: "Mandar Jagtap jagman.tech@xxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> Date: Sat, 11 Jun 2016 13:07:29 -0000 |
Thanks a lot Martin. It works well with your solution. On 10 Jun 2016 10:02 pm, "Martin Honnen martin.honnen@xxxxxx" < xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote: > On 10.06.2016 20:43, Mandar Jagtap jagman.tech@xxxxxxxxx wrote: > > I have a limitation to use MSXML 3.0 only as a XSLT processor but I >> want to get a URI or full path of source xml document that is being >> processed. Using Saxon or using XSLT 2.0, I can get it using >> document-uri(.). But, I don't seem to be able to find way to get it >> with limited MSXML 3.0 functions. >> >> Can somebody provide help on how to achieve it? >> > > You could insert some VBScript or JScript and read out the "url" property > the MSXML DOM exposes: > > <?xml version="1.0" encoding="UTF-8"?> > <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" > xmlns:msxsl="urn:schemas-microsoft-com:xslt" > xmlns:js="http://example.com/js" > exclude-result-prefixes="msxsl js" > version="1.0"> > > <msxsl:script language="JScript" implements-prefix="js"> > function documentUri(nodeSet) { > var node = nodeSet.item(0); > return node.nodeType === 9 ? node.url : > node.ownerDocument.url; > } > </msxsl:script> > > <xsl:output indent="yes"/> > > <xsl:template match="/"> > <results> > <result> > <xsl:value-of select="js:documentUri(/)"/> > </result> > <result> > <xsl:value-of select="js:documentUri(/*)"/> > </result> > </results> > > </xsl:template> > </xsl:stylesheet> > > > Note that in a quick test here I got a file URI returned for an input file > from the file system, only it had "_xml" appended at the end to the file > name.
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