- From: On Lee <onlee2000@hotmail.com>
- Date: Tue, 18 Aug 2009 15:32:23 -0700
- To: <public-sparql-dev@w3.org>
- Message-ID: <BAY104-DAV62EE3904FA068EDD350EBA0FF0@phx.gbl>
- Message-ID: <011801ca2053$c15e5640$441b02c0$@com>
http://danbri.org/foaf.rdf has the following nested info: <knows> <Person> <name>Aaron Swartz</name> <made> <rss:channel rdf:about="http://www.blogspace.com/rss/rss10"> <rss:title>RSS Info</rss:title> <rdfs:seeAlso rdf:resource="http://www.blogspace.com/rss/rss10"/> </rss:channel> </made> <rdfs:seeAlso rdf:resource="http://www.aaronsw.com/about.xrdf"/> </Person> </knows> http://sparql.org/sparql.html executed the following SPARQL: REFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#> PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#> PREFIX foaf: <http://xmlns.com/foaf/0.1/> SELECT * WHERE { <http://danbri.org/foaf.rdf#danbri> foaf:knows ?known . ?known ?predicate ?object . } It produced the first level info. For example, ?known = _:b19 ?predicate = <http://xmlns.com/foaf/0.1/made> ?object = <http://www.blogspace.com/rss/rss10> Based on the above info, I could then get the next level info by the following SPARQL: "SELECT * WHERE { <http://www.blogspace.com/rss/rss10> ?predicate ?object .}". How to write a single SPARQL so I could retrieve the nested level info even it has 10 levels deep? Thanks in advance for your help. Best, -- On Lee
Received on Tuesday, 18 August 2009 22:33:11 UTC