Subject: RE: [xsl] combining two variables to generate XPATH From: Mark Wilgus <mwilgus@xxxxxxxxx> Date: Tue, 17 Dec 2002 13:40:58 -0600 |
I believe this is not possible. XSLT doesn't allow you to use a variable for an XPath expression. See number 2 at the following list: http://www.dpawson.co.uk/xsl/sect2/nono.html > -----Original Message----- > From: bix xslt [mailto:bix_xslt@xxxxxxxxxxx] > Sent: Tuesday, December 17, 2002 1:32 PM > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > Subject: Re: [xsl] combining two variables to generate XPATH > > > Please note I made a slight correction. the $node that > appears within the > foreach statment just under the branch variable assignment > should be $branch > instead. This does not solve my problem - this is the > original problem. > Thanks. > > > >From: "bix xslt" <bix_xslt@xxxxxxxxxxx> > > <xsl:variable name="branch" > >select="concat($node,'/',$type,'/item')" /> > > <xsl:value-of select="$branch" /> > ><!-- <xsl:for-each select="$node"> > > becomes > > > <xsl:variable name="branch" > >select="concat($node,'/',$type,'/item')" /> > > <xsl:value-of select="$branch" /> > ><!-- <xsl:for-each select="$branch"> > > > _________________________________________________________________ > Help STOP SPAM with the new MSN 8 and get 2 months FREE* > http://join.msn.com/?page=features/junkmail > > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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