Pass the full paths to all files as a parameter to your transformation. Then
you can access them with the document() function.


Cheers,

Dimitre Novatchev [XML MVP],
FXSL developer, XML Insider,

http://fxsl.sourceforge.net/ -- the home of FXSL
Resume: http://fxsl.sf.net/DNovatchev/Resume/Res.html


"dSchwartz" <schwartz@c...> wrote in message
news:4ae1ece2.0403011455.495fe581@p......
> What I think I'm looking for is a way (in XSL stylesheet) to get the
> contents of a directory.  How can this be accomplished?
>
> more detailed:
>
> /root
>  index.aspx
>  xsl_style.xsl
>  /xml
>   newsletter_001.xml
>   newsletter_002.xml
>   newsletter_xxx.xml
>
>
> now i want to use xsl_style.xsl to pull two attributes from the root
> element of each and every file in the xml directory.  Something like
> this:
>
> for (each file in xml AS xmlFile)
>  <xsl:value-of select="document(xmlFile)//newsletter[@title]" />
>  <xsl:value-of select="document(xmlFile)//newsletter[@date]" />
>
>
> how do i create that loop?  I have to do this without a file that
> contains a list of the files in the xml directory, so i need to
> dynamically get the contents of the xml directory!
>
> Thanks for your time