Re: [xsl] for-each order

["Charly" <cohana at investacorp dot com>]

Nope,
That doesn't do it .


----- Original Message ----- 
From: "Long Zhao" <LZhao@firstgenetic.net>
To: <xsl-list@lists.mulberrytech.com>
Sent: Friday, December 14, 2001 9:34 AM
Subject: RE: [xsl] for-each order


> how about this
> 
> <xsl:variable name="position">
>    <xsl:for-each select="$path">
>       <xsl:value-of select="concat(@value, position)"/>,
>    </xsl:for-each>
> </xsl:variable>
> 
> Long
> 
> -----Original Message-----
> From: Charly [mailto:cohana@investacorp.com]
> Sent: Friday, December 14, 2001 2:35 PM
> To: xsl-list@lists.mulberrytech.com
> Subject: [xsl] for-each order
> 
> 
> Hello,
> Is there a way to do <xsl:for-each ..... and looping backwards.
> 
> My XML looks like 
> <histo>
>    <bar value="60" legend="Jan" />
>    <bar value="77" legend="Feb" />
>    <bar value="53" legend="Mar" />
>    <bar value="14" legend="Apr" />
> </histo>
> 
> and I want to store 14,53,77,60 in a variable .
> 
> 
> <xsl:variable name="position">
>    <xsl:for-each select="$path">
>       <xsl:value-of select="@value"/>,
>    </xsl:for-each>
> </xsl:variable>
> 
> 
> 
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
> 
> 
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
> 
> 


 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list