- From: Bob DuCharme <bob@snee.com>
- Date: Fri, 08 Oct 2010 10:24:02 -0400
- To: Michael Ransom <mransom@revelytix.com>
- CC: public-lod@w3.org
You can try this yourself with ARQ. Your query uses the property abc:hasMember, which makes sense in the context of each triple, but your data uses the the property abc:hasMembers, so the query won't find them. Once those were corrected in the data and the prefixes were declared, the following query did what you wanted: PREFIX abc: <http://www.example.com/ns/abc#> SELECT ?league (COUNT (?member) AS ?membercount) WHERE { ?league abc:hasMember ?member . } GROUP BY ?league Bob On 10/7/2010 12:02 PM, Michael Ransom wrote: > Hello All, > > I have a question about SPARQL 1.1 queries. > > > > > If have the following triples: > > :LeagueA abc:hasMembers :Alice, > :Bob, > :Carol . > :LeagueB abc:hasMembers :Dante, > :Edward. > > > > > If I want the following table of results: > > ?league ?membercount > LeagueA 3 > LeagueB 2 > > > > > Given the data and my desired results, will the following SPARQL 1.1 > query work? > > SELECT ?league (COUNT(?member) AS ?membercount) > WHERE { > SELECT ?league ?member > WHERE { > ?league abc:hasMember ?member . } GROUP BY ?league > } > > > > > Whether or not this query works, is there a way I can write this query > without a subquery? > > > > > Thank you.
Received on Friday, 8 October 2010 14:44:34 UTC