Subject: Re: [xsl] How can a template match elements created by xsl????? From: ChivaBaba@xxxxxxx Date: Thu, 16 May 2002 05:12:44 EDT |
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. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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