Advanced RDF and SPARQL

Introduction

We have Morrel, Max and Sasha being dogs, Sheeba and Query are cats, Picca is still a parrot, Fred and John are contacts. Fred claims that John is his friend. I changed the ontology to allow friendships between the animals too: Sasha claims that Morrel and Max are her friends. Sheeba claims Query is her friend. John bought Query. Fred being inspired by John decided to also get some pets: Morrel, Sasha and Sheeba.

Ontology

You can find the ontology here.

graph02.png As a .gaphor file

Let’s put this story in Turtle:

<test:Picca> a test:Parrot, test:Pet ;
        test:name "Picca" .

<test:Max> a test:Dog, test:Pet ;
        test:name "Max" .

<test:Morrel> a test:Dog, test:Pet ;
        test:name "Morrel" ;
        test:hasFriend <test:Max> .

<test:Sasha> a test:Dog, test:Pet ;
        test:name "Sasha" ;
        test:hasFriend <test:Morrel> ;
        test:hasFriend <test:Max> .

<test:Sheeba> a test:Cat, test:Pet ;
        test:name "Sheeba" ;
        test:hasFriend <test:Query> .

<test:Query> a test:Cat, test:Pet ;
        test:name "Query" .

<test:John> a test:Contact ;
        test:owns <test:Max> ;
        test:owns <test:Picca> ;
        test:owns <test:Query> ;
        test:name "John" .

<test:Fred> a test:Contact ;
        test:hasFriend <test:John> ;
        test:name "Fred" ;
        test:owns <test:Morrel> ;
        test:owns <test:Sasha> ;
        test:owns <test:Sheeba> .

Querytime!

Let’s first start with all friend relationships:

SELECT ?subject ?friend
WHERE { ?subject test:hasFriend ?friend }

  test:Morrel, test:Max
  test:Sasha, test:Morrel
  test:Sasha, test:Max
  test:Sheeba, test:Query
  test:Fred, test:John

Just counting these is pretty simple. In SPARQL all selectable fields must have a variable name, so we add the “as c” here.

SELECT COUNT (?friend) AS c
WHERE { ?subject test:hasFriend ?friend }

  5

We counted friend relationships, of course. Let’s say we want to count how many friends each subject has. This is a more interesting query than the previous one.

SELECT ?subject COUNT (?friend) AS c
WHERE { ?subject test:hasFriend ?friend }
GROUP BY ?subject

  test:Fred, 1
  test:Morrel, 1
  test:Sasha, 2
  test:Sheeba, 1

Actually, we’re only interested in the human friends:

SELECT ?subject COUNT (?friend) AS c
WHERE { ?subject test:hasFriend ?friend .
        ?friend a test:Contact
} GROUP BY ?subject

  test:Fred, 1

No no, we are only interested in friends that are either cats or dogs:

SELECT ?subject COUNT (?friend) AS c
WHERE { ?subject test:hasFriend ?friend .
       ?friend a ?type .
       FILTER ( ?type = test:Dog || ?type = test:Cat)
} GROUP BY ?subject"

  test:Morrel, 1
  test:Sasha, 2
  test:Sheeba, 1

Now we are only interested in friends that are either a cat or a dog, but whose name starts with a ‘S’.

SELECT ?subject COUNT (?friend) as c
WHERE { ?subject test:hasFriend ?friend ;
                 test:name ?n .
       ?friend a ?type .
       FILTER ( ?type = test:Dog || ?type = test:Cat) .
       FILTER REGEX (?n, '^S', 'i')
} GROUP BY ?subject

  test:Sasha, 2
  test:Sheeba, 1

Projects/Tracker/Documentation/AppDevelopersManual/AdvancedRDFandSPARQL (last edited 2013-11-25 12:54:22 by WilliamJonMcCann)