[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: for-each question
You're not specifying you want both documents to be generated. This is probably called the "xsl:for-each trap". Maybe you can try to do without the xsl for-each? xsl:value-of only gives you the first value! Try xsl:apply-templates here, or another xsl:for-each commando. If you really have to use xsl:for-each, why are you not specifying the xsl for-each as: <xsl:for-each select="/eventsCalendar/eventRecord[eventID=$neweventID]/listDoc/document"> for it is the content of your for-each that has to be outputted for each document, and not for each listdoc. I would suggest that you rewrite this fragment and try not to use xsl:for-each but xsl:apply-templates etc. Ronald At 12:34 PM 4/26/01 +0100, you wrote: I have the following xml: <eventsCalendar> <eventRecord> <eventID>1</eventID>> ---- <listDoc> <document href="C:\events1.doc">agenda.doc</document> <document href="C:\events2.doc">conference.doc</document> </listDoc> </eventRecord> ---- </eventsCalendar> XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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