Subject: [xsl] Return ancestor siblings based on Id From: "Adam J Knight" <adam@xxxxxxxxxxxxxxxxx> Date: Sat, 19 Feb 2005 01:16:58 +1000 |
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>
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