[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] correction: how to get new position() of a sorted resu
Hi, I apologize for the typo in my previous question. The original XSLT code is: <xsl:template match="/"> <xsl:apply-templates select="//article"> <xsl:sort data-type="number" order="descending" select="@date" /> </xsl:apply-templates> </xsl:template> <xsl:template match="article"> <xsl:if test="position() < 100"> ...... <!-- do processing here --> </xsl:if> </xsl:template> The original XML source file is: <?xml version="1.0" encoding="ISO-8859-1"?> <index section="Headlines"> <article filename="file1.xml"> <title>News 1 Title Text</title> <date>20001020</date> <ctprovider>Efe</ctprovider> </article> <article filename="file2.xml"> <title>News 2 Title Text</title> <date>20001113</date> <ctprovider>Eastside Journal</ctprovider> </article> <article filename="file3.xml"> <title>News 3 Title Text</title> <date>20001113</date> <ctprovider>Newsbytes News Network</ctprovider> </article> ............ <!-- more article elements here --> </index> The question is how to get the position() from a sorted result tree. In the above code, calling position() returns the ID of the original pre-sorted tree. Thanks, David XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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