Re: [xsl] position() in xsl:for-each

[Vidar Ramdal <vramdal@xxxxxxxxx>]

On Fri, 09 Jul 2004 15:06:17 +0300, George Cristian Bina <george@xxxxxxx> wrote:
>  > Ah. That explains everything.
>  >
>  > I think I'll copy the <object>s of the correct @type into another
>  > tree, and then use this tree to output the table correctly.
>  >
>  > That is, if there's no simpler solution :)
> 
> You can just get the following sibling that has the type1 type attribute:
> 
> <xsl:apply-templates select="following-sibling::*[@type='type1'][1]"/>

Yes, it's just that the real case is a bit more complicated than the
simplified example I posted here. I'll find a way though.
Thank you!

-- 
Vidar S. Ramdal