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.