xsl-list
[Top] [All Lists]

Re: position() in xsl:for-each

2004-07-09 05:06:17
> 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::*[(_at_)type='type1'][1]"/>

Best Regards,
George
-----------------------------------------------
George Cristian Bina
<oXygen/> XML Editor & XSLT Editor/Debugger
http://www.oxygenxml.com





<Prev in Thread] Current Thread [Next in Thread>