Re: [xsl] Numbering: position() ?

Subject: Re: [xsl] Numbering: position() ?
From: Sven Waibel <sven.waibel@xxxxxxxx>
Date: Wed, 10 Jan 2007 12:03:27 +0100
I forgot:

<xsl:template match="book">
    <xsl:apply-templates select="special | page"/>
</xsl:template>


<books>
  <book id="1">
 	<page>
 	<special>
 		<page>
 		<page>
 	</special>	
 	<page>
 	<page>
   </book>
</books>

My intention is to flaten down the tree structure to a list with<page>

Michael Kay wrote:
> 1. The value of position() within an xsl:template depends on the position of
> the node within the set of nodes selected by the select expression of the
> corresponding xsl:apply-templates, which you haven't shown us.
> 
> 2. You don't say what numbering you would consider correct. If you're
> looking for multi-level numbering, try <xsl:number level="multiple"/>. If
> you want single-level numbering, consider <xsl:number level="any"/>.
> 
> Michael Kay
> http://www.saxonica.com/
>  
> 
>> -----Original Message-----
>> From: Sven Waibel [mailto:sven.waibel@xxxxxxxx] 
>> Sent: 10 January 2007 10:41
>> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
>> Subject: [xsl] Numbering: position() ?
>>
>> Hi,
>>
>> i got following xml:
>>
>> <book id="1">
>> 	<page>
>> 	<special>
>> 		<page>
>> 		<page>
>> 	</special>	
>> 	<page>
>> 	<page>
>> </book>
>>
>> xslt:
>>
>>
>> <xsl:template match="page">
>> 	<xsl:value-of select="concat(concat(ancestor::book/$id, 
>> '.'), position())"/> </xsl:template>
>>
>> I got this numbering:
>> 1.1
>> 1.1
>> 1.2
>> 1.3
>> 1.4
>>
>> How can i achieve that the numbering is correct?
>>
>> Thanks
>> Sven
> 
> 
> 

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