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I'm not sure that I'm understanding you.
But if one of the groups is a <sequence>
it will be forcing an undesired ordering
on its members. And obviously none of them
can be a <choice>, so the only viable alternative
is to make them <all> groups. If you mean
that I could copy&paste the contents of the
common groups inside the others without
putting them between <all></all>, yeah,
you are right and that was what I wanted
to avoid, duplication.
Regards,
Carlos
On Thursday 31 March 2005 11:08, you wrote:
> I think you can just omit the xs:all from the groups you want to compose.
>
> Michael Kay
> http://www.saxonica.com/
>
> > -----Original Message-----
> > From: Carlos Pita [mailto:carlosjosepita@yahoo.com.ar]
> > Sent: 31 March 2005 14:09
> > To: xml-dev@lists.xml.org
> > Subject: [xml-dev] all group composition
> >
> > It seems as if this email has been rejected
> > so I'm reposting it. Sorry if you get it twice.
> > Regards,
> > Carlos
> >
> > Hi!
> >
> > Is it possible to compose unordered group models
> > using xml schema? I need to validate two elements
> > with a lot of common children but I prefer not
> > imposing any particular ordering on them. I would
> > like to factorize the 80% of common children into
> > an <all> group and then join them to the rest:
> >
> > <xs:group name="common">
> > <xs:all>
> > ....
> > </xs:all>
> > </xs:group>
> >
> > <xs:group name="group1">
> > <xs:all>
> > <xs:group ref="common"/>
> > ....
> > </xs:all>
> > </xs:group>
> >
> > <xs:group name="group2">
> > <xs:all>
> > <xs:group ref="common"/>
> > ....
> > </xs:all>
> > </xs:group>
> >
> > Of course I can't do this because <all> must be top
> > level in the group and moreover it can't contain <group>.
> >
> > Is it possible to achieve the factorization in another way?
> >
> > Thank you in advance.
> > Regards,
> > Carlos.
> >
> >
> >
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