Hello,
On 01/09/2017 02:11 PM, Lorenz Buehmann wrote:
> It was not my question, but again, the property path * operator is used
> in his query - and this can also mean 0 occurrences of the relation
> rdfs:subClassOf . Maybe I don't understand the meaning of * - I'll check
> the W3C recommendation.
I agree with Olivier (the other one) and Lorenz :
SELECT ?x ?rel ?y
WHERE {
?x rdf:type :mytype1 .
?y rdf:type :mytype2 .
?x ?rel ?y .
}
returns:
:A :myrel :X1
:B :myrel :X2
:C :myrel :X1
which is consistent with the graph
For ?x and ?y, adding propertypaths to also retrieve direct OR indirect
instances of mytype1 and mytype2 should potentially yield more values
for ?x and ?y (granted in his case exactly the same values as his graph
does not contain indirect instances), but not fewer values.
So either "?x rdf:type/(rdfs:subClassOf*) :mytype1" does not mean "the
direct or indirect instances of mytype1", or the following query should
return an empty set, according to Hugh and Kingsley (and in this case, I
still do not understand why it returns (A, myrel, X1) but not (B, myrel
X2) nor (C, myrel X1)):
SELECT ?x ?rel ?y
WHERE {
?x rdf:type/(rdfs:subClassOf*) :mytype1 .
?y rdf:type/(rdfs:subClassOf*) :mytype2 .
?x ?rel ?y .
}
returns only:
:A :myrel :X1
Can someone clarify this point?
Thank you all
olivier
------------------------------------------------------------------------------
Developer Access Program for Intel Xeon Phi Processors
Access to Intel Xeon Phi processor-based developer platforms.
With one year of Intel Parallel Studio XE.
Training and support from Colfax.
Order your platform today. http://sdm.link/xeonphi
_______________________________________________
Virtuoso-users mailing list
Virtuoso-users@lists.sourceforge.net
https://lists.sourceforge.net/lists/listinfo/virtuoso-users
