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RE: [xsl] Transforming XML to HTML Table


Subject: RE: [xsl] Transforming XML to HTML Table
From: "Nick Foster" <Nick.Foster@xxxxxxxxxxx>
Date: Wed, 29 Jan 2003 14:09:15 -0000

Oleg,

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. If they were always in the same order then there would be no problem in just looping through each and writing them out in the table. In each row I need to select the <sample> nodes in the same order that they appeared in the first row.

Thought: Is it possible to do a sort within each row so that they are all in the same order, alphabetical order perhaps?

Cheers,

Nick


-----Original Message-----
From: Oleg Tkachenko [mailto:olegt@xxxxxxxxxxxxx]
Sent: 29 January 2003 13:00
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: Re: [xsl] Transforming XML to HTML Table


Nick Foster wrote:

> but I only got the value of the first sample in all columns.  Is there a
> way of looping through the $columntitles/labels and selecting the sample
> where the label matches  in the current row?

Try
<xsl:for-each select="$columntitles">
	<TD>							
		<xsl:value-of select="$samples[label=current()/label]/displayval"/>
	</TD>
</xsl:for-each>

But what's order? You've got "sample 1" in first row and "Sample 1" in the 
second one. Is this a typo?

-- 
Oleg Tkachenko
eXperanto team
Multiconn Technologies, Israel


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