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

Subject: Re: [xsl] position() in xsl:for-each
From: George Cristian Bina <george@xxxxxxx>
Date: Fri, 09 Jul 2004 15:06:17 +0300
> 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]"/>

Best Regards,
George Cristian Bina
<oXygen/> XML Editor & XSLT Editor/Debugger

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