Subject: Re: [xsl] building path expressions around dynamic element node names From: David Carlisle <davidc@xxxxxxxxx> Date: Wed, 18 May 2005 13:49:27 +0100 |
If you test two node sets using = then you test the string value of each node in one set against the string value of each node in teh other set. <xsl:if test="../parent::node(.)=..//preceding-sibling::A/Class/text()">< ../parent::node(.) is a syntax error, nod() is not a function, and doesn't take an argument. you meant ../parent::node() which is the same as ../.. so the test will involve the string value of your grandparent, ie th econcatenation of all the text in all elements descended from your grandparent. is ../descendent-or-self::node()/preceding-sibling::A/Class/text( which is the same ../preceding-sibling::A/Class/text() | ..//A/Class/text() (The preceding-sibling axis is doing nothing when selecting descendents as the siblings are all selected anyway when selecting descendents, so preceding-sibling jjust selects them again, and then duplicates are removed as a node set is a set.) So this selects all the text nodes in A/Class elements that are preceding siblings or descendents of your parent. Each of these strings is compared to the strig value of ../.. It might be possible to construct a document where that test is true, but it seems unlikely except in very specially constructed cases. I'm not sure what you are trying to do. If you are on a Class element and want to select a child of Top that has the same name as string value of the current node then select="/Top/*[name()=currrent()] is all you need. 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 ________________________________________________________________________
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