Using Python rdflib: how to include literals in sparql queries?

问题

I can include URIs and variables in my queries, but I can't include literals in my queries.

Here, I have some code which successfully reads an RDF file, finds all the RDF triples with skos:prefLabels, counts them, and then identifies a couple of specific ones from a set of keywords:

import rdflib.plugins.sparql as sparql
import rdflib
import rdflib.graph as g

graph = g.Graph()
# Read the RDF file
graph.parse(
   'h:\......SKOSTaxonomy.rdf',
   format='xml')

# Build and execute the query
q = sparql.prepareQuery('SELECT ?s ?p ?o WHERE { ?s ?p ?o .}')
p = rdflib.URIRef("http://www.w3.org/2004/02/skos/core#prefLabel")
qres = graph.query(q, initBindings = {'p' : p})

print len(qres)

# Look for keywords among the results
keywords = set([u'Jackknifing', 'Technology-mapping', 'Something random'])

for (subj, pred, obj) in qres:
    if obj.value in keywords:
        print obj.value

As expected, this code prints:

2299
Jackknifing
Technology-mapping

since Jackknifing and Technology-mapping are prefLabels in the file.

What I really want to do is to construct and execute a Sparql query to look for each keyword in turn. But this is where I come unstuck, because I can't put a string into the query. I have tried this, for example:

o = rdflib.Literal(u'Jackknifing')
qres = graph.query(q, initBindings = {'p' : p, 'o' : o})

but qres is empty. I have also tried putting a literal explicitly into the query, e.g.

q = sparql.prepareQuery('SELECT ?s ?p WHERE { ?s ?p "Technology-mapping" .}')
qres = graph.query(q, initBindings = {'p' : p})

but that returns an empty result too.

How are literals included in a query?

回答1:

If the literals in your data have datatypes, or are strings with language tags, then a plain literal, that is, one without a datatype or language tag, injected into the query won't match.

The RDFLib docs on Literals show ways of creating literals with datatypes, but don't have an example of creating one with a language tag. However, the docs also have the source attached and the signature for Literal's __new__ is:

static __new__(lexical_or_value, lang=None, datatype=None, normalize=None)

Since the literal in your data has a language tag ('en'), you should to create your literal as

o = rdflib.Literal(u'Jackkifing',lang='en')

so that the language tag is associated with the literal.

来源：https://stackoverflow.com/questions/17144088/using-python-rdflib-how-to-include-literals-in-sparql-queries

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