RE: <xsl:sort> and <xsl:copy-of> in conjunction with o

Hi Jan,
	I looked over your xslt and nothing seems to be wrong there.  Could you post a sample of your xml?

Thanks,
Andy

-----Original Message-----
From: Stanger, Jan [mailto:jan.stanger@xxxxxxxx]
Sent: Wednesday, January 29, 2003 12:42 PM
To: 'xsl-list@xxxxxxxxxxxxxxxxxxxxxx'
Subject: RE:  <xsl:sort> and <xsl:copy-of> in conjunction with one
an other


I am still somewhat puzzled by below section of my code. I am trying to sort all DataRow nodes by up to five columns. Now the output actually looks unchanged from the input XML, as if the sorting would not have taken place at all. Is there anything obvious that I am missing? I have studied the online XSLT FAQ that I have been made aware of yesterday, but none of the sorting examples have helped me in this particular case.

I have checked all below variables (sortby, sortdatatype etc.) and they do contain correct values. Could it be that the copy-of instruction actually uses the original XML tree as opposed to the 'indexed' version of it after the sort?

Any help is appreciated.
Jan

-----Original Message-----

        <!-- Actual sorting of data rows -->            
        <xsl:for-each select="DataRow">
            <xsl:sort select="column[@name=$sortby0]" data-type="{$sortdatatype0}" order="{$sortorder0}" />
            <xsl:sort select="column[@name=$sortby1]" data-type="{$sortdatatype1}" order="{$sortorder1}" />
            <xsl:sort select="column[@name=$sortby2]" data-type="{$sortdatatype2}" order="{$sortorder2}" />
            <xsl:sort select="column[@name=$sortby3]" data-type="{$sortdatatype3}" order="{$sortorder3}" />
            <xsl:sort select="column[@name=$sortby4]" data-type="{$sortdatatype4}" order="{$sortorder4}" />
            <!-- Finally copy each DataRow node to output XML -->                
            <xsl:copy-of select="." />            
        </xsl:for-each>



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