Re: How can a template match elements created by xsl??

J.Pietschmann wrote:

>If this still doesn't fix your problem, check whether
> you have set a default namespace for the style sheet.

You was right again, it was a namespace-problem. as you see, I defined a default namespace:

++++++++++++++++++++++++++++++++++++++++++++++++++++++++
<xsl:stylesheet version="1.1"
        xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
        xmlns:exsl="http://exslt.org/common"
        extension-element-prefixes="exsl"
        xmlns="http://www.w3.org/TR/REC-html40">
++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Following your advice to check the variable with copy-of I realized that 'tool' belongs to the "html"-namespace.

I fixed the problem by creating " <tool xmlns=""> "  instead of " <tool> " and now the template matches what I want. 

But by checking the output I get now, I realized that the double sort you sent me some mails ago does not work right.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++

 <xsl:variable name="sw">
     <xsl:for-each select="exsl:node-set($tools)">
      <xsl:sort data-type="text" lang="en" select="@category"/>
      <xsl:sort data-type="text" lang="en" select="@name"/>
      <xsl:copy-of select="."/>
     </xsl:for-each>
    </xsl:variable>

++++++++++++++++++++++++++++++++++++++++++++++++++++++++

My tools in $tools are only sorted by category but not by name!!
What could the problem with this sort-instruction be????

Many Thanks and much respect,

Stefan.

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