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XSL question: position() strangeness

Ron Newman
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Given the following XML input:
� <test>
� <scan>1</scan>
� <scan>2</scan>
� <scan>3</scan>
� <scan>4</scan>
� <scan>5</scan>
� </test>
and the following XSL stylesheet:
� <xsl:stylesheet xmlns:xsl=""
� � � � � � � � version="1.0">
� <xslutput method="text"/>
� <xsl:template match="text()"/>
� <xsl:template match="test">
� � �<xsl:apply-templates/>
� </xsl:template>
� <xsl:template match="scan">
� � �content:<xsl:value-of select="."/>, position:<xsl:value-of select="position()"/>
� � �<xsl:if test="position() = last()"> I'M LAST!</xsl:if>
� </xsl:template>
� </xsl:stylesheet>
Why do I get this output?
� �content:1, position:2
� �content:2, position:4
� �content:3, position:6
� �content:4, position:8
� �content:5, position:10
It's acting as if there were other children of <test>, which there aren't.� It also doesn't seem to think the last child is really last.� This happens with both Xalan and Saxon.
If I change
� �<xsl:apply-templates/>
to either
� �<xsl:apply-templates select="*"/>
� �<xsl:apply-templates select="scan"/>
I get
� �content:1, position:1
� �content:2, position:2
� �content:3, position:3
� �content:4, position:4
� �content:5, position:5 I'M LAST!
which looks a lot more reasonable to me.
What's going on here?
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Hi Ron,
In a match template, the context is the whole input document, so the context position is equal to the numbered position of the node across all nodes (element, attribute, text node, comment, processing instruction)
The list of nodes is as follows:
1. test
2. scan
3. text()
4. scan
5. text()
If the template is applied without a select, you may use the preceding-sibling and following-sibling axis to achieve this result. The following template, for instance, would give the expected results:

An other way (cleaner) to do this is to use a for-each construct:

[ January 11, 2003: Message edited by: Beno�t d'Oncieu ]
It is sorta covered in the JavaRanch Style Guide.
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