Re: [xsl] Creating intermediate XML fragment

["alexandre bord" <alexandre.bord@xxxxxxxxxx>]

 Of course
i'm sorry not to have explained it .
The rules are
(0) @name = name1 = @ name2
(1) print foobar/@name if it is not ( in bar/@name1 or in bar/@name2 or 
in foo/@name)
(2) print bar/@name1 if it is not in foo/@name
(2') print bar/@name2 if it is not in foo/@name
(3) Don't print foo/@name

It's your second guess (i think)
Thanks once more for your help

alex

Peter Davis wrote:

> On Wednesday 18 September 2002 02:05, alexandre bord wrote:
>  
>
> > Hello !
> > Suppose my (piece of) xml looks like this :
> > <foobar name="fff" />
> > <foobar name="dddd" />
> > <bar name1="dddd" name2="xxx"/>
> > <bar name1="xxx" name2="yyy"
> > <foo name="yyy" />
> > <foo name="zzz" />
> >    
>
> No need for an intermediate fragment.  This is a simple grouping problem, just 
> with the additional complexity of the different meanings of the different 
> elements.
>
> I'm a little confused on your logic though.
>
>  
>
> > I'd like to output :
> > fff
> > dddd (once)
> > xxx (once)
> > (not 'yyy' nor 'zzz')
> >    
>
> Let me try and guess the rules:
>
> * For <foobar> elements, output @name if it is the first of 
> (foobar/@name|bar/@name1|bar/@name2) with the given value.
>
> * For <bar> elements, output @name1 and @name2 if they are the first of 
> (foobar/@name|bar/@name1|bar/@name2) with the given value and if the value is 
> not a value of a foo/@name.
>
> Or, another guess:
>
> * You want to treat foobar/@name, bar/@name1, and bar/@name2 as all the same, 
> and foo/@name is always not output?
>
> I'd be happy to help suggest a solution, but I don't want to waste my time if 
> my guesses are not correct.  Can you clarify a little bit?
>
>  



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