Re: SPARQL queries property or class from a schema

Thank you Mischa,

but in this way, how to decompose ?Property out? Could you please write a
full query?

Jitao

On Fri, Aug 12, 2011 at 19:20, Mischa Tuffield
<mischa.tuffield@garlik.com>wrote:

> Excuse the top-post.
>
> You can brutal about it in SPARQL 1.0 and use filter regex. Something like
> :
>
> . FILTER ( regex( str(?p), "/http:\/\/xmlns\.com\/foaf/") )
>
> It is kinda nasty, but will do the trick!
>
> Mischa
>
> Sent on the move
>
> On Aug 12, 2011, at 5:55 PM, Andy Seaborne <andy.seaborne@epimorphics.com>
> wrote:
>
> >
> >
> > On 12/08/11 17:36, Jitao Yang wrote:
> >> Dear all,
> >>
> >> suppose we have DATA:
> >>
> >> @prefix foaf: <http://xmlns.com/foaf/0.1/> .
> >>
> >> _:a  foaf:name "Johnny Lee Outlaw" .
> >> _:a  foaf:mbox <mailto:jlow@example.com <mailto:jlow@example.com>> .
> >> _:b  foaf:name "Peter Goodguy" .
> >> _:b  foaf:mbox <mailto:peter@example.org <mailto:peter@example.org>> .
> >> _:c  foaf:mbox <mailto:carol@example.org <mailto:carol@example.org>> .
> >>
> >>
> >> can we write a SPARQL query like:
> >>
> >> PREFIX foaf: <http://xmlns.com/foaf/0.1/>
> >> SELECT *?property*
> >>
> >> WHERE{
> >>             ?x  foaf:*?property* "Peter Goodguy".
> >> }
> >>
> >> if we can not, can we write an equivalent SPARQL query? Namely, can we
> >> decompose schema and Class(or Property) in SPARQL query?
> >>
> >> Thanks,
> >> Jitao
> >
> > SPARQL 1.1:
> >
> > PREFIX foaf:   <http://xmlns.com/foaf/0.1/>
> > SELECT ( substr( str(?p), strlen(str(foaf:))+1 )
> >          AS ?property)
> >
> > WHERE{
> >    ?x ?p  "Peter Goodguy".
> > }
> >
> >
> >
> > In SPARQL 1.0, you can't return anything not already in the graph : in
> SPARQL 1.1 you can do some manipulation of RDF terms.
> >
> > Because SUBSTR is fn:substring, indexes are 1-based.
> >
> >   Andy
> >
>
>
>