Subject: Re: [xsl] position() returns what? From: David Carlisle <davidc@xxxxxxxxx> Date: Fri, 24 Oct 2003 10:03:39 +0100 |
> But this outputs equal numbers: 2,4,6,8,10,... > I did expect that it would output 0,1,2,3,4,5 This is a FAQ (I'm sure itmustbe in teh faq file for this list) position() returns a number that is not a property of teh node (teh same node can have any number of position() values, for example if you select it with .. or . it always has position() =1. position() returns the position of the node in the current node list, and if that is the default selection for apply-templates this is child::node(0 which is all nodes including text nodes, so in your case the odd numbered nodes are all white space text nodes, if you added a template for them, <xsl:template match="text()"> <xsl:value-of select="position()"/> then you'd see it. change you apply tempaltes to use select="*" if you only want to process element children, or use xsl:strip-space to discard teh white space before processing begins. David -- http://www.dcarlisle.demon.co.uk/matthew ________________________________________________________________________ This e-mail has been scanned for all viruses by Star Internet. 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 ________________________________________________________________________ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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