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RE: [xsl] sorting based on a variable

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try data[number($column)]

Perhaps your $column variable is a string or a result tree fragment, not
a number. You didn't show us its declaration.

Michael Kay
Software AG
home: Michael.H.Kay@xxxxxxxxxxxx
work: Michael.Kay@xxxxxxxxxxxxxx 

> -----Original Message-----
> From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx 
> [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx] On Behalf Of 
> Kasper Nielsen
> Sent: 12 November 2002 14:34
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: [xsl] sorting based on a variable
> 
> 
> hi (another question)
> 
> lets say I want to some data for the $column data-column
> 
> <projects>
>   <project name="a">
>     <data> 1237</data>
>     <data> 1234</data>
>     <data> 1235</data>
>     <data> 1236</data>
>   </project>
>   <project name="b">
>     <data> 12</data>
>     <data> 41234</data>
>     <data> 51235</data>
>     <data> 71236</data>
>   </project>
>   <project name="c">
>     <data> 1</data>
>     <data> 41234</data>
>     <data> 51235</data>
>     <data> 71236</data>
>   </project>
> </projects>
> 
> so if $column=1 then it would sort them into {project 
> name="c", project name="b", project name="a"} however i've 
> tried <xsl:for-each select="projects/project">
>     <xsl:sort select="./data[$column]" data-type="number"/>
> but it doesn't work?
> however if just use something like
>     <xsl:sort select="./data[1]" data-type="number"/> it works fine
> 
> anyone can tell me what im doing wrong?
> 
> regards
>   Kasper Nielsen
> 
> 
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
> 


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