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Old February 3rd, 2006, 11:49 AM
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Default XSL Test

Given XML like this:

        <page id="5004" desc="Site" sort="5004">
            <page id="5242" desc="Careers" sort="5242">
                <page id="350231838" desc="Careers Update" sort="350231838">
                    <page id="350231839" desc="September 2005" sort="350231839" />
        <page id="350231839" />
I want to transform the pages (please ignore the badly formed XML) into a series of unordered lists, but I don't want to output any elements if they appear in the <hidden> section. My current test before I output the ul is xsl:if "child::page", but this does not properly handle the case where the current element has children, but they all appear in the hidden section. How would I ask (essentially), "If this element has at least one child element 'page' that does not show up in the hidden list"?

Old February 3rd, 2006, 12:35 PM
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Think you need the not function:
<xsl:stylesheet version="1.0"
  <xsl:output method="html"/>

  <xsl:template match="/">
          <xsl:apply-templates select="nav/tree/page[not(@id = /nav/hidden/page/@id)]" />

  <xsl:template match="page">
      <xsl:value-of select="@desc"/>
      <xsl:if test="page[not(@id = /nav/hidden/page/@id)]">
          <xsl:apply-templates select="page[not(@id = /nav/hidden/page/@id)]"/>
This assumes that when you hit a hidden item it has no more visible children.


Joe (Microsoft MVP - XML)
Old February 3rd, 2006, 02:49 PM
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Perfect. Thanks. I didn't know how to reference elements in that way.

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