What SPARQL query will return the data from the example?

Question

In my RDF store I have the following triples

For schema:

ex:animal rdf:type rdf:Class;
    rdfs:subClassof ex:Package_animal.

ex:height rdf:type rdf:Property;
    rdfs:label "height";
    rdfs:domain ex:animal.

ex:age rdf:type rdf:Property;
    rdfs:label "age";
    rdfs:domain ex:animal.

ex:NumberOfLegs rdf:type rdf:Property;
    rdfs:label "NumberOfLegs";
    rdfs:domain ex:animal.

For instances:

ex:dog rdf:type ex:animal;
    ex:height "10";
    ex:age "2";
    ex:NumberOfLegs "4".

ex:cat rdf:type ex:animal;
    ex:height "10";
    ex:age "3";
    ex:NumberOfLegs "4".

I want to:

Get properties of class ex:animal

ex:height 
ex:age
ex:NumberOfLegs

Get the instance ex:dog of class ex:animal with all of its properties

ex:dog rdf:type ex:animal
ex:dog ex:height "10"
ex:dog ex:age "2"
ex:dog ex:NumberOfLegs "4"

The SPARQL I thought was:

SELECT ?s
WHERE
{
   ?p rdf:type rdf:Property.
   ?s ?p ?y.
   ?s rdf:type ex:dog.
}

what does not work with your query modulo using proper (prefixed) URIs which I suppose is just here because you consider it a sketch? — UninformedUser, Nov 30 '22 at 13:08
Your data is also a bit vague - besides those weird trailing ----A - you should consider proper datatypes for your number values, I would not use a string to model the number of legs or the age, use an integer here — UninformedUser, Nov 30 '22 at 13:10
I edited your question a bit more, so check if that's what you wanted to ask. — RedCrusaderJr, Dec 04 '22 at 00:45

score 1 · Accepted Answer · answered Dec 04 '22 at 01:29

This query

SELECT ?propertyName ?value
WHERE
{
    ex:dog ?propertyName ?value.
}

will give you this result

propertyName    value
rdf:type        ex:animal
ex:height       "10"
ex:age          "2"
ex:NumberOfLegs "4"

You can bound ex:dog URI if you want it in the result

SELECT ?dog ?propertyName ?value
WHERE
{
    BIND(ex:dog as ?dog).
    ?dog rdf:type ex:animal.
    ?dog ?propertyName ?value.
}

which will yield

dog     propertyName     value
ex:dog  rdf:type         ex:animal
ex:dog  ex:height        "10"
ex:dog  ex:age           "2"
ex:dog  ex:NumberOfLegs  "4"

If you want to grab all properties of ex:animal you'd need to change your data. I edited your question so it contains the data shown below. For scheme knowledge:

   ex:animal rdf:type rdf:Class;
      rdfs:subClassof ex:Package_animal.

   ex:height rdf:type rdf:Property;
      rdfs:label "height";
      rdfs:domain ex:animal.

   ex:age rdf:type rdf:Property;
      rdfs:label "age";
      rdfs:domain ex:animal.

   ex:NumberOfLegs rdf:type rdf:Property;
      rdfs:label "NumberOfLegs";
      rdfs:domain ex:animal.

For instances knowledge:

   ex:dog rdf:type ex:animal;
      ex:height "10";
      ex:age "2";
      ex:NumberOfLegs "4".

   ex:cat rdf:type ex:animal;
      ex:height "10";
      ex:age "3";
      ex:NumberOfLegs "4".

Then you can use this SPARQL query:

SELECT ?dog ?animalPropertyName ?value
WHERE
{
    ?animalPropertyName rdfs:domain ex:animal.
    
    BIND(ex:dog as ?dog).  
    ?dog ?animalPropertyName ?value.
}

which will yield

dog     animalPropertyName  value
ex:dog  ex:height           "10"
ex:dog  ex:age              "2"
ex:dog  ex:NumberOfLegs     "4"

What SPARQL query will return the data from the example?

1 Answers1