Re: [xsl] document function and counting

[Francis Norton <francis@xxxxxxxxxxx>]

All I can say is it was late and I was tired and it looked plausible...
in other words I was being lazy and I knew it...

Thanks for straightening that out!

Francis.

Trevor Nash wrote:
> 
> >How about something like this?
> >
> ><untested>
> >...
> ><xsl:for-each select="document('foo-1.xml')//item |
> >document('foo-2.xml')//item" >
> >       <xsl:value-of select="position()" />
> >       <xsl:copy-of select="." />
> ></xsl:for-each>
> >...
> ></untested>
> >
> >Francis.
> >
> A case where testing would not guarantee a bulletproof answer!
> 
> The trouble is, for-each processes the nodes in 'document order'.
> >From the spec:
> "There are no constraints on how the implementation orders documents
> other than that it must do so consistently: an implementation must
> always use the same order for the same set of documents."
> 
> In other words a processor might decide to give you all the items from
> foo-2 first.  I think the spec even lets the same processor give you
> different answers on different runs, depending on how strongly you
> take 'consistent'.  It could certainly be the case that it might base
> the order on the URIs of the documents.
> 
> Pity, it is a nice idea.
> 
> Regards,
> Trevor Nash
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