"Randy Oxentenko" <randy(_at_)oxentenko(_dot_)com> wrote in message
news:001301c301da$5eef5f70$7d2d0fcf(_at_)randyoxentenko(_dot_)(_dot_)(_dot_)
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
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list