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

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Subject: Re: 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
-----------------------------------------------
George Cristian Bina
<oXygen/> XML Editor & XSLT Editor/Debugger
http://www.oxygenxml.com

Current Thread

Re: position() in xsl:for-each, (continued)
- George Cristian Bina - Fri, 09 Jul 2004 12:03:13 +0300
  - Vidar Ramdal - Fri, 9 Jul 2004 11:26:10 +0200
    - David Carlisle - Fri, 9 Jul 2004 10:47:38 +0100
    - Vidar Ramdal - Fri, 9 Jul 2004 13:25:42 +0200
    - George Cristian Bina - Fri, 09 Jul 2004 15:06:17 +0300 <=
    - Vidar Ramdal - Fri, 9 Jul 2004 14:14:13 +0200
- David Carlisle - Fri, 9 Jul 2004 10:10:02 +0100
  - Vidar Ramdal - Fri, 9 Jul 2004 11:34:20 +0200

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