[xsl] xalan:nodeset or xsl:copy-of problem

Subject: [xsl] xalan:nodeset or xsl:copy-of problem
From: Ming <myu@xxxxxxxx>
Date: Tue, 23 Apr 2002 16:24:43 -0600
Hi,

I'm having problems with my stylesheet and after debugging, I found that
the problem may lie in the xalan:nodeset (or xsl:copy-of)part.

Here is part of my xml file:
<ztitle>
   <title db="dbName">title of the article</title>
</ztitle>

Here comes my stylesheet and the problem is explained in the comments:

 <xsl:template name="setup_db">
    <xsl:param name="field"/>
    <xsl:param name="db_list"/>
    <xsl:param name="full_path"/>


    <xsl:variable name="nodetree">
      <xsl:for-each select="$full_path[contains($dbs_searched,@db)]">
<!-- $dbs_searched variable is passed from a different stylesheet -->
        <xsl:copy-of select="."/>
      </xsl:for-each>
    </xsl:variable>

<!-- This statement does print out the attribute value of field $field,
which means that $full_path is a valid path and does contain @db
attribute -->
    <xsl:value-of select="$full_path/@db" />

<!-- This statement does print out the value of the field $field, which
means that $nodetree does have something there and does contain the
right $field value -->
    <xsl:value-of select="xalan:nodeset($nodetree)/*[name()=$field]" />

<!-- But, this statement doesn't give me the attribute value of the
field $field. It returned nothing. -->

    <xsl:value-of select="xalan:nodeset($nodetree)/*[name()=$field]/@db"
/>

<!-- some other codes -->

</xsl:template>

So, the problem seems to be that either the copy-of command doesn't copy
the whole tag or the xalan:nodeset wasn't be able to do the job.

Thanks a lot for your help.

Ming


 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list


Current Thread