Subject: xsl:sort From: Dave Pawson <daveP@xxxxxxxxxxxxxxxxxxxxxxx> Date: Sun, 06 Aug 2000 11:34:36 +0100 |
I thought I understood sorting :-| With the following xml <?xml version="1.0" encoding="utf-8"?> <faqindex> <functions/> <functions> <pair> <word>document()</word> <file>N169.html</file> </pair> </functions> <functions> <pair> <word>order</word> <file>N795.html</file> </pair> </functions> <functions> <pair> <word>position()</word> <file>N961.html</file> </pair> </functions> I want to sort on select = pair/word. Does a sort use the key *only* for sorting, i.e. can I keep the function and its children together durin the sort? <xsl:template match="faqindex"> <H2>Index for XSLT FAQ Website</H2> <!-- Sort the functions --> <xsl:variable name="fns"> <xsl:for-each select="functions[.!='']"> <xsl:sort data-type="text" select="pair/word"/> </xsl:for-each> </xsl:variable> <!-- now convert that to a node-set and step through it --> <xsl:for-each select="xt:node-set($fns)">* <xsl:value-of select="child::*/text()"/> <br /> </xsl:for-each> </xsl:template> this is giving me zilch output, and I can't see the logic. I'm hoping that the fns variable will hold the node-list of all function elements, sorted by 'word' child, and include the file child... or at least thats the way I thought it worked. any help appreciated. TIA DaveP XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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