Sparqlwrapper: Its not returning anything but the query is working with dbpedia.org/sparql

Question

I have also added the snippet below. This query will return the name of the company uri, its name and parent company. It is working with DBpedia.org/sparql but not with sparqlwrapper(not returning anything; )

query1 =  """
    PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
    PREFIX dbo: <http://dbpedia.org/ontology/>
    PREFIX dbp: <http://dbpedia.org/property/>
    PREFIX db: <http://dbpedia.org/resource/>
    select distinct ?iri ?name ?parent (concat('[',group_concat(distinct ?location;separator=','),']') as ?location)
    {
        ?iri a dbo:Company ;
            rdfs:label ?label ;
            foaf:name ?name
            OPTIONAL { ?iri dbo:parentCompany ?parent. 
            filter (!isBlank(?parent))     }
            OPTIONAL { ?iri dbo:location ?location.}
            filter(regex(?name, "\\btata steel\\b","i" )) .
            filter(regex(?label, "\\btata steel\\b","i" ))
    }
    GROUP BY ?iri ?name ?parent 
"""
def get_country_description(query1):
    sparql = SPARQLWrapper("http://dbpedia.org/sparql")
    sparql.setReturnFormat('json')

    sparql.setQuery(query1)  # the previous query as a literal string

    return sparql.query().convert()

Answer 1

get rid of \\b in your regex expression, and it works! (i dont know why it is needed as tata steel is a whole word)

Also you need to pass query1! to your function! and use sparql.setReturnFormat('json') not JSON,

cheers