Subject: Re: [xsl] Comparing node for identity using union From: António Mota <amsmota@xxxxxxxxx> Date: Wed, 19 Jan 2005 19:08:29 +0000 |
Neverthless, just for the record, please note Sarissa.setXslParameter(xslDoc, "pos", pos+""); xml.transformNodeToObject(xslDoc, xmlResult); ----> Works OK Sarissa.setXslParameter(xslDoc, "pos", pos); xml.transformNodeToObject(xslDoc, xmlResult); ----> DOES NOT processor.setParameter(null, "pos", pos); var xmlResult = processor.transformToDocument(xmlDoc); ----> Works OK processor.setParameter(null, "pos", pos+""); var xmlResult = processor.transformToDocument(xmlDoc); ----> DOES NOT Works OK = $pos is recognized as a number so (//Menu)[$pos] gives the Menu as position $pos DOES NOT = $pos is NOT recognized as a number so (//Menu)[$pos] gives ALL the Menu nodes, as just //Menu does. On Wed, 19 Jan 2005 18:10:20 -0000, Michael Kay <mike@xxxxxxxxxxxx> wrote: > > > > > > The $pos is defined like > > > > > > > > <xsl:param name="pos" select="0"/> > > > > > > That's certainly a number > > > > I think it's only a number as long as it doesn't get overridden - as > > soon as you pass in the value a parameter it becomes a string. > > Of course: I missed that. > > Shows how useful it is to declare your types in XSLT 2.0. > > Michael Kay > http://www.saxonica.com/
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