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Re: SPARQL queries property or class from a schema

From: Mischa Tuffield <mischa.tuffield@garlik.com>
Date: Fri, 12 Aug 2011 18:20:23 +0100
Message-Id: <6B666ABF-64DB-4685-9A5D-526CD124F094@garlik.com>
Cc: "semantic-web@w3.org" <semantic-web@w3.org>
To: Andy Seaborne <andy.seaborne@epimorphics.com>
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
> 
Received on Friday, 12 August 2011 17:22:14 UTC

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