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Re: [xsl] Return ancestor siblings based on Id
Subject: Re: [xsl] Return ancestor siblings based on Id From: Mukul Gandhi <mukul_gandhi@xxxxxxxxx> Date: Fri, 18 Feb 2005 07:34:00 -0800 (PST) |
I think union (|) operator is the answer.. //tree_node[@id=9]/ancestor::tree_node | //tree_node[@id=9]/preceding-sibling::* | //tree_node[@id=9] | //tree_node[@id=9]/following-sibling::* Probably, this should also work (seems more efficient also) - //tree_node[@id=9]/(ancestor::tree_node | preceding-sibling::* | self::* | following-sibling::*) Regards, Mukul --- Adam J Knight <adam@xxxxxxxxxxxxxxxxx> wrote: > Hi guys, > > Another chapter in this tragedy. > > Question: > > The expression > '//tree_node[@id=9]/ancestor::tree_node' returns a > nodeset > containing node with id(7). I have verified this > with xpath visualizer. > > How do I return the ALL siblings of this node also. > > The result being > Node(id=7) > Node(id=8) : sibling of Node(id=7) > Node(id=9) : sibling of Node(id=7) > Node(id=14) : sibling of Node(id=7) > > HELP APPRECIATED!!!! > > ALL TO0 FAMILIAR XML STRUCTURE: > <tree> > <tree_node id="7" value="Test Level One A"> > <tree_node id="8" value="Test Level Two A"/> > <tree_node id="9" value="Test Level Two B"> > <tree_node id="11" value="Test Level Three > B"/> > <tree_node id="10" value="Test Level Three A"> > <tree_node id="12" value="Test Level Four > A"/> > <tree_node id="13" value="Test Level Four > B"/> > </tree_node> > </tree_node> > <tree_node id="14" value="Test Level Two C"/> > </tree_node> > </tree> __________________________________ Do you Yahoo!? Yahoo! Mail - Helps protect you from nasty viruses. http://promotions.yahoo.com/new_mail
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Re: [xsl] Return ancestor siblings , António Mota | Thread | [xsl] java extension element and xs, Keith Lynch |
Re: [xsl] Return ancestor siblings , António Mota | Date | [xsl] Correction: Return Ancestor N, Adam J Knight |
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