Subject: RE: parent position() From: Melle Zegel <melle.zegel@xxxxxxxxxxxx> Date: Wed, 11 Oct 2000 14:21:33 +0200 |
position(.) perhaps -----Oorspronkelijk bericht----- Van: David Carlisle [mailto:davidc@xxxxxxxxx] Verzonden: woensdag 11 oktober 2000 14:01 Aan: xsl-list@xxxxxxxxxxxxxxxx Onderwerp: Re: parent position() I believe position() is applied to the current context node. I want to know the value of position() of the parent. How can I do This? Because I'm dealing with a while lot of recursiveness in my documents I cannot, give position of the parent and than process the children. I HAVE to process the children and only then I want to know the position() of the parent. Does this make any sense at all?? I more or less wanted to use something like: <xsl:value-of select="../position()"/> Allright I know this is wrong, but it's probably the only way to describe my problem. you probably don't want position() at all. position() gives the position in the current node list (ie the nodes selected by the currently active apply-templates or for-each. So if you selct the parent with .. the position() is always 1 as th enode list selected just has one node. Probably you want something like <xsl:for-each select=".."> <xsl:number/> </xsl:for-each> David 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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