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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] RE: replacing an output block
 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@xxxxxxxxxxx] > Sent: 27 June 2005 11:23 > To: XSL List > Subject: 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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