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Re: [AccessingRdfLists] relationship with property paths?

From: Chimezie Ogbuji <ogbujic@ccf.org>
Date: Tue, 31 Mar 2009 18:25:49 -0400
To: "Lee Feigenbaum" <lee@thefigtrees.net>, "SPARQL Working Group" <public-rdf-dawg@w3.org>
Message-ID: <C5F810AD.90E1%ogbujic@ccf.org>
During today's teleconference the issue about whether or not property path
support is sufficient for supporting access to RDF lists came up.  I happen
to think property path support is generally useful for more than just
meeting this particular requirement.  For the caveat below, assuming the
scenario where you don't know what the start of the list is (which I would
imagine is rare), wouldn't you be able to achieve a result without
duplicates with the following query?:

PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
SELECT * 
{ 
  UNSAID { [] rdf:rest ?list }
  ?list rdf:rest*/rdf:first ?member
} 

I.e., so the path matching *begins* with _:b1?

Note: Sorry for the use of two feature requests, but w/out UNSAID, it would
be cumbersome to write :)

-- Chimezie

Andy wrote:
> PREFIX  rdf:    <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
> SELECT * { ?list rdf:rest*/rdf:first ?member }
> Because if the data is:
> ---- Data:
> ("A" "B" "C") .
> Which is in triples:
> ---- Same data:
> _:b1      rdf:first  "A" .
> _:b1      rdf:rest  _:b2 .
> _:b2      rdf:first  "B" .
> _:b2      rdf:rest  _:b3 .
> _:b3      rdf:first  "C" .
> _:b3      rdf:rest  rdf:nil .
> ----

> The results are:
> -----------------
> | list | member |
> =================
> | _:x0 | "B"    |
> | _:x0 | "C"    |
> | _:x1 | "A"    |
> | _:x1 | "B"    |
> | _:x1 | "C"    |
> | _:x2 | "C"    |
> -----------------
> If ?list is bound it would work.



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Received on Tuesday, 31 March 2009 22:26:58 GMT

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