[xsl] Function position()

Subject: [xsl] Function position()
From: "Partho Paul" <uk4u@xxxxxxxxxxxxxxxxxxx>
Date: Thu, 01 Feb 2001 12:33:37 +0100
I have the following xml-structure:

                                           /            \
                                        /                  \
                             Paragraph              Paragraph
                            /  |  \    \                      /   |   \
                        /      |     \     \                    etc.
                    /          |        \     \              
                /              |          \      \           
            Text   StretchText  Text   StretchText

Now I need in each StretchText-Node the number of itself.
That means what is the number of the StretchText - node if I look at 
the whole tree. So I can count  ALL Stretchtext-nodes from 1 to n.

If I use 'position()' I only get the node number of Stretchtext from the node 'Paragraph',
So each Stretchtext child is countet from 1 to m in each Paragraph,
but I need to count in the whole tree.

I used the following:

<xsl:template match"Chapter">
 <xsl:apply-template  select="Paragraph"/>

<xsl:template match"Paragraph">
 <xsl:value-of select="Text"/>
 <xsl:apply-template  select="StretchText"/>

<xsl:template match"StretchText">
 <xsl:value-of select="position()"/>        ---> THIS doesn't work in my case.
Isn't it possible to do like the following: can't I tell the function 'position()' to count
the StretchText-nodes from the root node and not from Paragraph????

Things like <xsl:value-of select="Chapter//StretchText[position()]"/> does not work.

Thank you.

Partho Paul                     email: uk4u@xxxxxxxxxxxxxxxxxxx
Computer science student        or:    paul@xxxxxxxxx
Homepage:   http://www.partho.de
Durga Puja: http://www.uni-karlsruhe.de/~uk4u/durga.html

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