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RE: replacing an output block

2005-06-27 04:28:32
You're thinking procedurally. There's no "earlier" or "later" in XSLT -
order of execution is undefined.

Sounds as if you need a two phase stylesheet. Phase 1 computes a result tree
and puts this in a variable. Phase 2 considers whether a better result is
possible: if so, it applies a transformation to the first result (replacing
the relevant subtree with a better version and copying everything else
across unchanged); if not, it simply returns the result of the first phase.

Michael Kay
http://www.saxonica.com/


-----Original Message-----
From: Rahil [mailto:qamar_rahil(_at_)yahoo(_dot_)co(_dot_)uk] 
Sent: 27 June 2005 11:23
To: XSL List
Subject: [xsl] replacing an output block

Hi

I want to replace a block of an HTML TABLE output in case 
better results 
are obtained later. So I have the result in 'first.html' of some 
processing in 'first.xsl'.

first.html -- output

<HTML><BODY>
    <TABLE>
          <TR>
                <TD>Want to replace this block with new output</TD>
                <TD>Some other results from first.xsl</TD>
           </TR>
    </TABLE>
</BODY></HTML>

The block shown above has been copied as such in second.xsl using 
<xsl:copy> and <xsl:copy-of>.

However as a result of processin some other block a better match is 
obtained for the TR/TD[1] above and I would like to replace this new 
finding with the old one. Given that I can easily locate the position 
where this TR/TD[1] occurs, how do I overwrite the earlier output ?

Would appreciate any help or suggestions.

Thanks
Rahil




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