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Subject: RE: [xsl] Assigning multiple xml's file names into variable name using for-each possible? From: "Michael Kay" <mike@xxxxxxxxxxxx> Date: Tue, 18 Jul 2006 14:59:14 +0100 |
> So, what I was trying to do is to make a seperate xml file
> listing all the xml file to be processed. e.g: xmlFileList.xml
>
> <ContentGroup name="Arts">
> <ContentFile>S207.xml</ContentFile>
> <ContentFile>S103_2.xml</ContentFile>
> <ContentFile>DD03.xml</ContentFile>
> .... etc etc
> </ContentGroup>
>
>
> And in the xsl file, trying to loop thro to get each file
> name and process it like:
>
> <xsl:template match="/"> <!-- of xmlFileList.xml -->
> <xsl:for-each select="//ContentFile">
> <xsl:variable name="fileName" select="node()"/>
> <xsl:variable name="currentFile"
> select="document('{$fileName}')"/>
>
You just want
select="document($fileName)"
Curly braces are only used when an XPath expression is embedded in attribute
text, never to embed one XPath expression inside another.
Michael Kay
http://www.saxonica.com/
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