RE: xpath question
HI Florian, You can use the local-name() function to solve this problem. It returns the (unqualified) name of the context element. In these circumstances you would usually select with a wildcard, and then use local-name() in a predicate to filter it down. Something like: <xsl:template match="column"> <xsl:for-each select="/root/group/*[local-name()=current()/@name]"> <!-- Do whatever --> </xsl:for-each> </xsl:template> The above selects the actual fruity element. If you want the <group> change the xpath in the for-each to "/root/group[*[local-name()=current()/@name]]". Cheers, Stuart > -----Original Message----- > From: florian [mailto:csshsh@xxxxxxxxxxxxxxx] > Sent: 27 March 2003 17:08 > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > Subject: xpath question > > > > hi! > > i have a document xml doc like this: > > <root> > <group> > <apple>bla</apple> > <orange>bla</orange> > </group> > <group> > <apple>bla</apple> > <orange>bla</orange> > </group> > > <order> > <column name="orange" /> > <column name="apple" /> > </order> > </root> > > i would like to do the following: im going though all the column nodes > and would like to access the group nodes where the column > attribute name > and the group node name match up. > > basically i can just not think of a way to do that in xpath.. anybody > got an tip? how can i say in xpath that it should get the node with > the name in @name and not just access the name attribute of a group > node.. > > thanks alot! > > ciao! > florian > > > 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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