Subject: Re: [xsl] XSLQUERRY From: "cutlass" <cutlass@xxxxxxxxxxx> Date: Wed, 27 Feb 2002 12:45:53 -0000 |
----- Original Message ----- From: "XSL Developer" <vsd18@xxxxxxxxxxxxxx> > > dear xsl members, > i have a problem. here it goes.. > xml structure: > > <root> > <schedule category="title" type = "title2"> > title2 > </schedule> > <schedule category="title" type = " title1"> > title1 > </schedule> > <schedule category="title" type = "title3"> > episode > </schedule> > <schedule category="description" type="desc1"> > desc1 > </schedule> > </root> this should do the trick, though i havent tested. xml file ---------------------------------------- <?xml version="1.0" ?> <root> <schedule category="title" type="title2"> title2 </schedule> <schedule category="title" type="title1"> title1 </schedule> <schedule category="title" type="title3"> episode </schedule> <schedule category="description" type="desc1"> desc1 </schedule> </root> xsl file ---------------------------------------- <?xml version="1.0"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template match="root"> <xsl:for-each select="schedule[@category='title' and @type='title1'] "> <xsl:value-of select="."/> </xsl:for-each> </xsl:template> </xsl:stylesheet> > and i want to get the text where category = title and type = > title1 btw, i would avoid stating something like XSLQuery, in a subject line, as it could be misconstrued to be XMLQuery, a completely different thing chow, jim fuller XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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