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Re: position() in xsl:for-each

2004-07-09 05:14:13
On Fri, 09 Jul 2004 15:06:17 +0300, George Cristian Bina 
<george(_at_)sync(_dot_)ro> 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::*[(_at_)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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