Subject: Re: position() in xsl:for-each
From: Vidar Ramdal <vramdal@xxxxxxxxx>
Date: Fri, 9 Jul 2004 14:14:13 +0200
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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
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