Re: XPATH - Finding similar/conflicting nodes

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">

<xsl:output indent="yes"/>

The equal case  is fairly easy I think:

<xsl:template match="/">
equal:
<xsl:copy-of select="/SCHED/Schedule/Day/Squad/event[@id='2']
 [@timebeg = ../preceding-sibling::Squad/event[@id='2']/@timebeg]
"/>
</xsl:template>

</xsl:stylesheet>

produces

$ saxon squad.xml squad.xsl
<?xml version="1.0" encoding="utf-8"?>
equal:
<event id="2" timebeg="960" timend="1020"/>



> Now I have an XPATH question.

I don't think that you can find the overlapping cases in a single Xpath 1
statement (although I could be proved wrong) You could do it in Xpath2
or using an XSLT loop construct.

David


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