How to get query result separated by columns?

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I need to make a query in dbpedia and from this query return name, gender and other information of a particular singer. I am able to do this, but my code returns the whole tuple, I wish I could separate the information by column. Can someone help me? Follows the code:

<?php
    require_once('sparqllib.php');
    $db = sparql_connect('http://dbpedia.org/sparql');
    $query = "  PREFIX owl: <http://www.w3.org/2002/07/owl#>
                PREFIX xsd: <http://www.w3.org/2001/XMLSchema#>
                PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
                PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
                PREFIX foaf: <http://xmlns.com/foaf/0.1/>
                PREFIX dc: <http://purl.org/dc/elements/1.1/>
                PREFIX : <http://dbpedia.org/resource/>
                PREFIX dbpedia2: <http://dbpedia.org/property/>
                PREFIX dbpedia: <http://dbpedia.org/>
                PREFIX skos: <http://www.w3.org/2004/02/skos/core#>
                PREFIX owl: <http://dbpedia.org/ontology/>
                PREFIX rsc: <http://dbpedia.org/resource/>
                    SELECT ?name ?hometown ?origin ?genre ?bandMember ?currentMembers ?associatedMusicalArtist
                    WHERE { 
                      rsc:Arctic_Monkeys dbpedia2:name ?name.
                      rsc:Arctic_Monkeys owl:hometown ?hometown.
                      rsc:Arctic_Monkeys dbpedia2:origin ?origin .
                      rsc:Arctic_Monkeys dbpedia2:currentMembers ?currentMembers .
                      rsc:Arctic_Monkeys owl:genre ?genre .
                      rsc:Arctic_Monkeys owl:bandMember ?bandMember .
                      rsc:Arctic_Monkeys owl:associatedMusicalArtist ?associatedMusicalArtist .

                    FILTER ((LANG(?name) = 'en') AND (LANG(?origin) = 'en')).
                    }";

    $result = sparql_query($query);
    $fields = sparql_field_array($result);
    while($row = sparql_fetch_array($result))
    {
      foreach($fields as $field)
      {
        print"$row[$field] \n";
      }
    }
?>
php sparql

  • You mean this? echo $Row['name']." ( ".$Row['origin']." ) <br/>";

  • That’s right, Dalton, that’s right! But I did the test by swapping "print"$Row[$field] n";" for "print"$Row['name'] n";" and it didn’t work... =/ gave the error "Parse error: syntax error, Unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING"

  • try it like this: print $row[$field]."<br>\n"; (with or without the <br>, depending on the desired result). Alternatively, so: print $field.': '.$row[$field]."<br>\n";

2 answers

1

Simply print the column by its respective index.

print ( $row['name']."\n".$row['origin'] );

I don’t know on Linux, but on Windows for \n break the line, it needs the function nl2br.

Staying:

print nl2br( ( $row['name']."\n".$row['origin'] ) );

-1

How about trying with the Freebase.com?

{ "id": null, "name": "Arctic Monkeys", "type": "/music/Artist" }

Freebase also uses Wikipedia data and all requests are in JSON, so it’s easy to ask for any specific information.

  • If the Dbpedia result is an "array"=>"associative" just add foreach($Fields as $key=>$value)

  • sorry Gaucho, but I didn’t understand the solution... could you give this example using the name field that is returned from the query? thanks!

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