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Function position()
- To: "xsl" <xsl-list at lists dot mulberrytech dot com>
- Subject: [xsl] Function position()
- From: "Partho Paul" <uk4u at rz dot uni-karlsruhe dot de>
- Date: Thu, 01 Feb 2001 12:33:37 +0100
- Reply-To: xsl-list at lists dot mulberrytech dot com
I have the following xml-structure:
Chapter
/ \
/ \
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>
<xsl:template match"Paragraph">
<xsl:value-of select="Text"/>
<xsl:apply-template select="StretchText"/>
</xsl:template>
<xsl:template match"StretchText">
<xsl:value-of select="position()"/> ---> THIS doesn't work in my case.
</xsl:template>
---------------------------------------------------------------------------------------------------------------
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
_______________________________________________________________________
Partho Paul email: uk4u@rz.uni-karlsruhe.de
Computer science student or: paul@partho.de
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