Subject: Re: [xsl] Automatically generate xpath From: David Carlisle <davidc@xxxxxxxxx> Date: Tue, 31 Jan 2006 00:48:20 GMT |
>You showed me how to use union for two xpaths but how would I remove a node > from a xpath? you dont want to remove nodes from a xpath (which is an expression) but from a node set (which is a value). So you are asking for set difference: given two node sets $a and $b generate a set of nodes in $a but not $b. As I think someone showed earlier, in XPath2 this is $a except $b in XPath1 set difference isn't a standard operator but it's available using the standard idiom: $a[count(.|$b)!=count($b)] David ________________________________________________________________________ This e-mail has been scanned for all viruses by Star. The service is powered by MessageLabs. For more information on a proactive anti-virus service working around the clock, around the globe, visit: http://www.star.net.uk ________________________________________________________________________
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
Re: [xsl] Automatically generate xp, Liron | Thread | Re: [xsl] Automatically generate xp, Liron |
RE: [xsl] ordering nodes, bokluk | Date | Re: [xsl] Automatically generate xp, Liron |
Month |