RE: counting the element number in a recursive tree
If position() returns even numbers, it's because you are counting text nodes as well as elements. The answer is to count only the elements, which will happen if you use <xsl:for-each select="*"> or <xsl:apply-templates select="*"> rather than using select="node()". Dividing position() by 2 is wrong, because the whitespace text nodes won't always be there. Alternatively, use xsl:number. Michael Kay http://www.saxonica.com/ > -----Original Message----- > From: Mohsen Saboorian [mailto:mohsens@xxxxxxxxx] > Sent: 26 June 2006 12:06 > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > Subject: counting the element number in a recursive tree > > Hi, > I'm confused with counting the current element number with a > recursive xml structure. Here is a sample: > <node> > <node> > </node> > <node> > <node> > <node> > </node> > </node> > </node> > <node> > </node> > </node> > > = = = = = = = = = > Since position() returns even numbers (I think is counts also > #text elements), I used position div 2, but unfortunately > deeper nodes also have problems. > > <xsl:template match="node"> > <xsl:value-of select="position() div 2" /> > > <xsl:apply-templates /> > </xsl:template> > > What souled I do to find out the correct number of node I'm > currently on. > > Thanks in advance.
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