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RE: [xsl] XPath: all elements with only non-parent children with identical style attr
Subject: RE: [xsl] XPath: all elements with only non-parent children with identical style attr From: "Chris Bayes" <chris@xxxxxxxxxxx> Date: Wed, 12 Dec 2001 00:27:46 -0000 |
Tobi, Not sure exactly what you want but I guess this will do <xsl:template match="foo"> <xsl:if test="(count(*) = count(*[@style = current()/*[1]/@style])) and not(*[node()])"> <xsl:copy-of select="." /> </xsl:if> </xsl:template> Using this xml <root> <foo> <bar style="baz"/> <blah style="baz"/> <blam style="baz"/> <blam style="baz" /> </foo> <foo> <bar style="baz"/> <blah style="baz"/> <blam style="baz"/> <blam style="baz"> <z /> </blam> </foo> <foo> <bar style="baz"/> <blah style="baz"/> <blam style="baz"/> <blam style="baz"> <!----> </blam> </foo> <foo> <bar style="bish"/> <blah style="baz"/> <blam style="baz"/> <blam style="baz"> <z /> </blam> </foo> <foo> <bar style="baf"/> <blah style="baz"/> <blam style="baz"/> <blam style="baz"/> </foo> </root> Produces <foo> <bar style="baz"/> <blah style="baz"/> <blam style="baz"/> <blam style="baz"/> </foo> i.e. only the first one. Ciao Chris XML/XSL Portal http://www.bayes.co.uk/xml XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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