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RE: Transforming XML to HTML Table

2003-01-29 07:09:15
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(_at_)multiconn(_dot_)com]
Sent: 29 January 2003 13:00
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
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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