Re: [xsl] xml include

[David Carlisle <davidc@xxxxxxxxx>]

   Hi,
       I have two xml files, one of them is included in other:

   <?xml version="1.0"?>
   <document xmlns:xinclude="http://www.w3.org/1999/XML/xinclude";>
     <xinclude:include href="xml.xml"/>
   </document>

   And I use this xsl to transform:
   <?xml version="1.0"?>
   <xsl:stylesheet version='1.0'
   xmlns:xsl='http://www.w3.org/1999/XSL/Transform' >
   <xsl:output method="html" />
   <xsl:template match="/">
   <H1><xsl:value-of select="//title"/></H1>
   <H2><xsl:value-of select="//author"/></H2>
   </xsl:template>
   </xsl:stylesheet>

   (xml.xml contains title and author tags)

   The problem is that xsl cannot read title and author tags ... ?

The input to XSL is (logically) a tree. If your XML parser natively
supported xml:include (do any?) and "expanded includes" the way that
it expands entities, then the tree would have <author> as a descendent
of the root node of the input document and your stylesheet would work.

As it is, xml:include is probably just seen as a general element by all
process concerned so you have to follow the link yourself.


   <?xml version="1.0"?>
   <xsl:stylesheet version='1.0'
   xmlns:xsl='http://www.w3.org/1999/XSL/Transform'
      xmlns:xinclude="http://www.w3.org/1999/XML/xinclude";>

   <xsl:output method="html" />
   <xsl:template match="/">
   <xsl:variable name="x" select="document(document/xinclude:include/@href)"/>
   <H1><xsl:value-of select="$x//title"/></H1>
   <H2><xsl:value-of select="$x//author"/></H2>
   </xsl:template>
   </xsl:stylesheet> >
 

David

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