RE: [xsl] determine node position

Subject: RE: [xsl] determine node position
From: "Michael Kay" <mhkay@xxxxxxxxxxxx>
Date: Tue, 10 Apr 2001 13:46:16 +0100
> I tried already anything like this:
> <xsl:when test="News/NewsBlock[@LABEL = '4']"><xsl:value-of
> select="position()">
> </xsl:when>
> but in this case i get always node position 1.

Your extract doesn't tell us what the current node list is. position()
returns the position of the node in the current node list, which is
established by a call on xsl:for each or xsl:apply-templates.

> how to determine the position towards the whole node-set ?
> <week>
> <item ID="monday"/>
> <item ID="thuesday"/>
> <item ID="wednesday"/>
> <item ID="thursday"/>
> <item ID="friday"/>
> <item ID="saterday"/>
> <item ID="sunday"/>
> </week>
> I want to now for example on which node position <item
> ID="thursday"/> is.

If you want the position of the node among all the children of its parent
node, use <xsl:number/> or count(preceding-sibling::item)+1.

Mike Kay
Software AG

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