SPARQL查询以检索所有对象和属性

Question

Using the Wine ontology , I want to create SPARQL query so I can retrieve all wines and their properties like the table below - consider that I don't know the properties' names a priori .

vin                  | rdf:type     | vin:hasMaker      |  vin:hasSugar   | ...
==========================================================================  ...
GaryFarrellMerlot    |  vin:Merlot  | vin:Elyse         |  vin:Dry        | ...
--------------------------------------------------------------------------
ElyseZinfandel       |  vin:Elyse   | vin:GaryFarrell   |  vin:Dry        | ...
...

Can someone give me a hint?

-- EDIT

It is not possible to have the query result in the format I mentioned, but I can have it this way:

vin                |   property     |  value
=================================================
GaryFarrellMerlot  |   rdf:type     |  vin:Merlot
-------------------------------------------------
GaryFarrellMerlot  |   rdf:hasMaker |  vin:Elyse
-------------------------------------------------
...

With this select (thanks cygri ):

SELECT DISTINCT ?wine ?property ?value
WHERE { 
       ?o1 a ?class .
       ?wine a ?o1 .
       ?wine ?property ?value .
}

This will take wines instead of type of wines (Merlot -> GaryFarrellMerlot). The only problem is it takes wine but also Winery, Regions, Flavours, etc and I wanted only wines and its properties. Besides, there's no such attribute vin:Merlot rdfs:subClassOf vin:Wine. Any hint?

Answer 1

You can't. You need to know beforehand what columns you want in your query result.

Listing all properties of wines isn't very hard of course:

SELECT DISTINCT ?property
WHERE { 
    ?wine a vin:Wine .
    ?wine ?property ?value .
}

Then you need to write some code that creates the final SPARQL query from the list of results.