Question

我尝试通过此Sparql查询获取结果绑定。通过此Sparql入口点：http://digitale.bncf.firenze.sbn.it/openrdf-workbench/repositories/NS_03_2014/query

我有错误：“TypeError：query（）至少需要2个参数（给定1个）

谢谢!!!

@app.route('/caricaArgomento/<type>', methods=['GET'])
def getArgomento(type):
    #sparql = SPARQLUpdateStore("http://digitale.bncf.firenze.sbn.it/openrdf-      workbench/repositories/NS_03_2014/query")
    sparql=SPARQLUpdateStore("http://digitale.bncf.firenze.sbn.it/openrdf-workbench/repositories/NS_03_2014/query")
    sparql.setQuery("""
    PREFIX dc:<http://purl.org/dc/elements/1.1/>
    PREFIX rdfs:<http://www.w3.org/2000/01/rdf-schema#>
    PREFIX nsogi:<http://prefix.cc/nsogi>
    PREFIX rdf:<http://www.w3.org/1999/02/22-rdf-syntax-ns#>
    PREFIX skos:<http://www.w3.org/2004/02/skos/core#>
    PREFIX dcterms:<http://purl.org/dc/terms/>
        SELECT ?risultato
        WHERE {
         ?item a skos:Concept .
         ?item skos:prefLabel ?risultato .
         filter regex(?risultato, """+type+""", "i")
        } ORDER BY  ?risultato
         """)

     #FILTER regex(str(?aConcept), "http://thes.bncf.firenze.sbn.it/", "i").}

    sparql.setReturnFormat(JSON)
    results = sparql.query().convert()
    results = results['results']['bindings']
    results = json.dumps(results)

return results

digitale.bncf.firenze.sbn.it digitale.bncf.firenze.sbn.it

Answer 1

查看相关的documentation sparql.query()...需要查询参数。

sparql=SPARQLUpdateStore("http://digitale.bncf.firenze.sbn.it/openrdf-workbench/repositories/NS_03_2014/query")

sparql.setReturnFormat(JSON)

query = """
    PREFIX dc:<http://purl.org/dc/elements/1.1/>
    PREFIX rdfs:<http://www.w3.org/2000/01/rdf-schema#>
    PREFIX nsogi:<http://prefix.cc/nsogi>
    PREFIX rdf:<http://www.w3.org/1999/02/22-rdf-syntax-ns#>
    PREFIX skos:<http://www.w3.org/2004/02/skos/core#>
    PREFIX dcterms:<http://purl.org/dc/terms/>
    SELECT ?risultato
    WHERE {
     ?item a skos:Concept .
     ?item skos:prefLabel ?risultato .
     filter regex(?risultato, """+type+""", "i")
    } ORDER BY  ?risultato
"""

results = sparql.query(query).convert()

我无法确定sparql.setQuery(...)的目的是什么，但很明显，这不是你想要的。

编辑：

查看源setQuery(...)实际上是供内部使用（例如，人们编写子类），而不是常规的api用户。它拉出查询表单并记录查询文本。 query(...)在内部调用它。

results = results ['results'] ['bindings'] Flask错误

1 个答案: