Question

利用Wine ontology , 我想制造SPARQL问询,因此我可以检索all wines和their nature,如下表,我认为我不知道这些财产名称a priori。

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

谁能给我一个 h?

页: 1

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
-------------------------------------------------
...

选择(cygri):

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

这将使用 wine而不使用 wine(Merlot-> GaryFarmMerlot)。唯一的问题是 wine,但也包括 Wine、区域、 Fl等,我只想 wine及其财产。此外,还有no <>/strong>这种属性vin:Merlot rdfs:subClassOf vin:Wine.。任何 h?

Answer 1

你们能够这样做。你们需要事先了解你在问询结果中想要哪一栏。

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.

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