How to query dbpedia resource ontology 'wikiPageExternalLink'

Question

Using sparql\sparqlwrapper in python, how will I be able to query for the values of a certain dbpedia resource? For example, how will I be able to get the dbpedia-owl:wikiPageExternalLink values of http://dbpedia.org/page/Asturias? Here's a simple example on how will I be able to query for the rdfs:label of Asturias. But I don't know how to modify the query/query parameters to get values of property/ontology other than those included on rdfs schema. Here's the sample:

from SPARQLWrapper import SPARQLWrapper, JSON, XML, N3, RDF
sparql = SPARQLWrapper("http://dbpedia.org/sparql")
sparql.setQuery("""
    PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
    SELECT ?label
    WHERE { <http://dbpedia.org/resource/Asturias> rdfs:label ?label }
""")
print '\n\n*** JSON Example'
sparql.setReturnFormat(JSON)
results = sparql.query().convert()
for result in results["results"]["bindings"]:
    print result["label"]["value"]

Hoping to receive feedback. Thanks in advance!

Answer 1

Not sure where you're stuck—this is really easy:

SELECT ?label
WHERE { <http://dbpedia.org/resource/Asturias>
            dbpedia-owl:wikiPageExternalLink ?label }

Usually you need to declare the namespace prefixes like rdfs: or dbpedia-owl: if you want to use them in the query, but on the DBpedia endpoint this works even without. If you want, you can declare them anyways:

PREFIX dbpedia-owl: <http://dbpedia.org/ontology/>
SELECT ?label
WHERE { <http://dbpedia.org/resource/Asturias>
            dbpedia-owl:wikiPageExternalLink ?label }

You can find out the full URI corresponding to the prefix by going to http://dbpedia.org/sparql and clicking on “Namespace Prefixes” near the top right corner.

If you want to rename the variable (for example, from ?label to ?link) then do it like this:

PREFIX dbpedia-owl: <http://dbpedia.org/ontology/>
SELECT ?link
WHERE { <http://dbpedia.org/resource/Asturias>
            dbpedia-owl:wikiPageExternalLink ?link }

and you also have to change "label" to "link" in the Python code that gets the value out of the JSON result.