Re: [xsl] Transforming XML to HTML Table

Subject: Re: [xsl] Transforming XML to HTML Table
From: Oleg Tkachenko <olegt@xxxxxxxxxxxxx>
Date: Wed, 29 Jan 2003 16:28:35 +0200
Nick Foster wrote:

No it's not a typo, the <sample> nodes may appear in any order within each
<row> node so that the order in the first row might be Sample 1, Sample 2,
Sample 3 but in the next row it could be Sample 2, Sample 3, Sample 1.
You don't understand. In your example first row contains of "sample 1", but second one contains "Sample 1" (note case). That's why label=current()/label doesn't work. If you change "sample 1" to "Sample 1" it works.

Thought: Is it possible to do a sort within each row so that they are all
in the same order, alphabetical order perhaps?
Sure. Something like
<xsl:for-each select="samples">
	<xsl:sort data-type="text" select="label"/>
	<TD>							
		<xsl:value-of select="displayval"/>
	</TD>
</xsl:for-each>

--
Oleg Tkachenko
eXperanto team
Multiconn Technologies, Israel


XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list



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