RE: [xsl] Grouping Question

["Michael Kay" <mike@xxxxxxxxxxxx>]

Just change

<xsl:apply-templates select="."/>

to

<xsl:apply-templates select="current-group()"/> 

Actually, since you're not making changes, I think you could do

<xsl:copy-of select="current-group()"/> 

One other point: the href attribute of xsl:result-document is supposed to be
a URI, not a Windows filename.

Michael Kay
http://www.saxonica.com/


> -----Original Message-----
> From: Danny Leblanc [mailto:leblancd@xxxxxxxxxxxxxxxxxxx] 
> Sent: 11 June 2007 19:49
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: [xsl] Grouping Question
> 
> Hello everyone,
> 
>   Let say I have an XML file that looks like this
> 
> <root>
>   <L1>Data</L1>
>   <L1>Data</L1>
>   <L1>Data</L1>
>   <L1>Data</L1>
>   <L1>Data</L1>
>   <L1>Data</L1>
>   <L1>Data</L1>
>   <L1>Data</L1>
> </root>
> 
>   What I want to do is split this into multiple files each 
> time a new L1 is found. The following code is what I use. 
> (The code for this is more or less generic except for the XPATH).
> 
> <?xml version="1.0" encoding="utf-8"?>
> <xsl:stylesheet version="2.0" 
> xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
> <xsl:output method="xml" version="1.0" encoding="utf-8" 
> indent="yes"/> 
>     <xsl:template match="/">
>         <xsl:for-each select="root/L1">
>          <xsl:result-document 
> href="C:\\out\\{format-number(position(),'000000000')}.xml">
>           <reportrun>
>           <batch>
>               <xsl:apply-templates select="."/>
>           </batch>
>           </reportrun>
>          </xsl:result-document>
>         </xsl:for-each>  
>     </xsl:template>
>   <xsl:template match="@*|node()">
>     <xsl:copy>
>       <xsl:apply-templates select="@*|node()"/>
>     </xsl:copy>
>   </xsl:template>
> </xsl:stylesheet>
> 
>   This works A1, just the way I want it to. Now what I would 
> like to do is add an option that would allow the user to do 
> the same split but they could choose how many "L1" would go 
> into the output file. For example, right now the above case 
> creates 8 files, one per L1. I would like to output 4 files 
> that would each contain 2 L1 nodes.
> 
>   I tried something like this 
> 
> <?xml version="1.0" encoding="utf-8"?>
> <xsl:stylesheet version="2.0" 
> xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
> <xsl:output method="xml" version="1.0" encoding="utf-8" 
> indent="yes"/> 
>     <xsl:template match="/">
>         <xsl:for-each-group select="root/L1" 
> group-by="L1[position() mod 5 = 0]">
>          <xsl:result-document 
> href="C:\\out\\{format-number(position(),'000000000')}.xml">
>           <reportrun>
>           <batch>
>               <xsl:apply-templates select="."/>
>           </batch>
>           </reportrun>
>          </xsl:result-document>
>         </xsl:for-each-group>  
>     </xsl:template>
>   <xsl:template match="@*|node()">
>     <xsl:copy>
>       <xsl:apply-templates select="@*|node()"/>
>     </xsl:copy>
>   </xsl:template>
> </xsl:stylesheet>
>   
>   Which did not work. Any insights as to what I would have to 
> change to get this going would be appreciated.
> 
> Thank you in advance.