"Randy Oxentenko" <randy@xxxxxxxxxxxxx> wrote in message
news:001301c301da$5eef5f70$7d2d0fcf@xxxxxxxxxxxxxxxxx
> I am a novice with XSLT.  My question is with respect to the child of
> <xsl:for-each> that is <xsl:sort>.  Is it possible to use a result
returned
> from <xsl:call-template> as the sort key?

This would be possible in XSLT 2.0 using xsl:function

For now you can use a generic sort template, which accepts a "compare"
template as a parameter. See for example:

http://www.biglist.com/lists/xsl-list/archives/200303/msg00007.html


=====
Cheers,

Dimitre Novatchev.
http://fxsl.sourceforge.net/ -- the home of FXSL




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