In 2.0:

select="document(document('index.xml')/protest/package/name/concat(.,
'.xml'))">

In 1.0:

<xsl:for-each select="document('index.xml')/protest/package/name">
  <xsl:for-each select="document(concat(., '.xml'))">

    ....


Michael Kay
http://www.saxonica.com/


> -----Original Message-----
> From: chun ji [mailto:cji_work@xxxxxxxxx] 
> Sent: 04 December 2006 23:27
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: [xsl] new bie question 
> 
> I have an index.xml file, which reads as: 
> "
> -<?xml version="1.0" encoding="ISO-8859-1" ?> 
> -<?xml-stylesheet href="rules.xsl" type="text/xsl"?> -<protest>
>     -<package>
>            -<name>billing</name>
>       -</package>
>      -<package>
>            -<name>timecard</name>
>       -</package>
> -</protest>
> ". 
> There is a group of child xml files that associated with each 
> "/protest/package/name", such as "billing.xml", "timecard.xml". 
> 
> Now I have soming in my XSL file that tries to open all these 
> CHILD xml files, " <xsl:for-each 
> select="document(document('index.xml')/protest/package/name)">
> .
> </xsl:for-each>
> "
> which fails, because I did not give ".xml" for each
> child xml.   
> 
> Does someone know the format to do it ? 
> 
> 
> thanks a lot 
> 
> 
> cji 
> 
> 
>  
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