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RE: Transforming XML to HTML Table

2003-01-29 07:52:07
Oleg,

Thanks for that, I will give it a try.

Someone else has suggested that I change 

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

to 

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

and that seems to have worked but I will try the sorting as well to see which 
gives me the best performance.

Thanks for your help,

Nick

-----Original Message-----
From: Oleg Tkachenko [mailto:olegt(_at_)multiconn(_dot_)com]
Sent: 29 January 2003 14:29
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] Transforming XML to HTML Table


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


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