Formatting SPARQL Results in dotNetRDF

Question

Hi,

I have a SPARQL Query that I am executing through dotNetRDF. The query is given as below:

Object results = store.ExecuteQuery("PREFIX pre:<http://www.owl-ontologies.com/LRBrain_Simple.owl#>" + "SELECT ?name ?Characteristics ?Behaviour " + "WHERE { ?Person pre:hasCharacteristics ?Characteristics. ?Person pre:name ?name. ?Characteristics pre:showsBehaviour ?Behaviour} " + "ORDER BY ?name");

When I display the result in Console using the following:

SparqlResultSet rset = (SparqlResultSet)results; foreach (SparqlResult result in rset) { Console.WriteLine(result.ToString()); }

The result is displayed as follows:

?Characteristics = http://www.owl-ontologies.com/LRBrain_Simple.owl#Intutive , ?Behaviour = http://www.owl-ontologies.com/LRBrain_Simple.owl#RightBrained , ?name = A^^http://www.w3.org/2001/XMLSchema#string ?Characteristics = http://www.owl-ontologies.com/LRBrain_Simple.owl#Linear , ?Behaviour = http://www.owl-ontologies.com/LRBrain_Simple.owl#LeftBrained , ?name = A^^http://www.w3.org/2001/XMLSchema#string

I want to format this output as, say, a tabular format like:

Name | Characteristics | Behaviour A Intutive RightBrained A Linear LeftBrained

or atleast like:

Name = A Characteristics = Intutive Behaviour = RightBrained

Name = A Characteristics = Linear Behaviour = LeftBrained

I went through the link: How can you format SPARQL Query Results in dotnetrdf? and tried working out the solution provided, but the line if (r.HasVariable(var) && r[var] != null) shows an error as there is no HasVariable function for SparqlResult.

So I modified the code, leaving out the r.HasVariable and the following code:

SparqlResultSet rset = (SparqlResultSet)results; foreach (SparqlResult r in rset) { foreach (String var in r.Variables) { if (r[var] != null) { INode n = r[var]; String output; switch (n.NodeType) { case NodeType.Literal: output = ((ILiteralNode)n).Value; Console.Write(output); break;

        case NodeType.Uri:
        output = ((IUriNode)n).ToString();
        Console.Write(output);
        break;

        default:
        output = n.ToString();
        Console.Write(output);
        break;
    }
  }
  Console.WriteLine();
}

}

The above code gives me the answer as:

http://www.owl-ontologies.com/LRBrain_Simple.owl#Intutive http://www.owl-ontologies.com/LRBrain_Simple.owl#RightBrained A http://www.owl-ontologies.com/LRBrain_Simple.owl#Linear http://www.owl-ontologies.com/LRBrain_Simple.owl#LeftBrained A

How can I split only the word Intutive from http://www.owl-ontologies.com/LRBrain_Simple.owl#Intutive and also format it in the way mentioned above?

Accepted Answer

You need to edit the code that deals with URI nodes from the example code you have reproduced, currently you have the following:

case NodeType.Uri:
        output = ((IUriNode)n).ToString();
        Console.Write(output);
        break;

This is just going to give you the full URI string, what you likely want to do is get the URI and add some actual logic e.g.

case NodeType.URI:
    Uri u = ((IUriNode)n).Uri;
    if (!u.Fragment.Equals(String.Empty))
    {
        //u.Fragment is the fragment ID, the #blah bit of the URI
        //You might need to add .Substring(1) here to avoid including the # 
        //since I can't remember off the top of my head if .Net includes it
        output = u.Fragment;
    }
    else
    {
       //Otherwise use the full URI
       output = u.AbsoluteUri;
    }
    Console.WriteLine(output);
    break;

If you want to make more complex formatting decisions you need to add more of your own logic.

Formatting SPARQL Results in dotNetRDF

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