SPARQLWrapper 1.5.2

SPARQL Wrapper

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LICENSE TYPE:
W3C License 
USER RATING:
UNRATED
  0.0/5
DEVELOPED BY:
Ivan Herman, Sergio Fernández, Carlos ...
HOMEPAGE:
sparql-wrapper.sourceforge.net
CATEGORY:
ROOT \ Database \ Database APIs
SPARQLWrapper is a SPARQL endpoint interface to Python. It is a wrapper around a SPARQL service. It helps in creating the query URI and, possibly, convert the result into a more manageable format.

The SPARQLWrapper package is licensed under W3C license.

Here you have an example of how to use the library in your python program:

from SPARQLWrapper import SPARQLWrapper, JSON

sparql = SPARQLWrapper("http://dbpedia.org/sparql")
sparql.setQuery("""
 PREFIX rdfs:
 SELECT ?label
 WHERE { rdfs:label ?label }
""")
sparql.setReturnFormat(JSON)
results = sparql.query().convert()

for result in results["results"]["bindings"]:
 print result["label"]["value"]

Last updated on August 30th, 2012

requirements

#SPARQL wrapper #Python module #SPARQL interface #SPARQL #wrapper #interface #Python

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