RE: [xsl] position() of parent node

Subject: RE: [xsl] position() of parent node
From: "Michael Kay" <mhk@xxxxxxxxx>
Date: Mon, 17 Nov 2003 22:16:52 -0000
> 
> 
> This should be an easy one, but it is eluding me.
> 
> XML:
> <manual>
> 	<chapter>
> 		<textLink>para one</textLink>
> 		<textLink>para two</textLink>
> 		<imageLink>image one</imageLink>
> 		<imageLink>image two</imageLink>
> 	</chapter>
> </manual>
> 
> XSLT:
> <xsl:for-each select="chapter">
> 		<xsl:for-each select="textLink">
> 			//do some stuff
> 		</xsl:for-each>
> 		<xsl:for-each select="imageLink">
> 			<xsl:if test="chapter[position() != 21]">
> 				//do your thing
> 			</xsl:if>
> 		</xsl:for-each>
> 	</xsl:for-each>

You are testing whether the imageLink has a child chapter whose position
is not 21. It doesn't have any child chapters, and if it did have any,
it would certainly have one whose position was not 21.

position() doesn't test the position of a node in the source tree, it
tests the position of a node in the sequence currently being processed
(e.g. by for-each or apply-templates).

So you want:

 <xsl:for-each select="chapter">
 		<xsl:for-each select="textLink">
 			//do some stuff
		</xsl:for-each>
            <xsl:if test="position() != 21">
 		  <xsl:for-each select="imageLink">
 			//do your thing
 		  </xsl:for-each>
            </xsl:if>
 	</xsl:for-each>

Michael Kay



> 
> What I am getting at, is that I want to output all text links 
> (easy enough), and all images for chapters unless the chapter 
> is 21 (which happens also to be last, which I tried <xsl:if 
> test="chapter[position() != last()]">)
> 
> Position has given me troubles in the past.
> 
> Robert Ogden
> IETM Developer
> Navy Programs
> (763) 572-7121
> 
>  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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