- From: Steve Harris <steve.harris@garlik.com>
- Date: Fri, 3 Aug 2012 10:24:27 +0100
- To: James Rothering <jr@jamesrothering.com>
- Cc: semantic-web@w3.org
- Message-Id: <BAFBE145-7EEC-4063-A834-6EB700195D98@garlik.com>
On 2 Aug 2012, at 21:54, James Rothering wrote:
> Is is possible to implement an inheritance hierarchy in a SPARQL query?
> In other words, is it possible to select all from a parent type, but
> overriding from a child type wherever an entry exists?
>
> I.e., given this data:
> ex:parent001 rdf:type ex:"Parent"
> ex:widgetName ex:"test"
> ex:hasChild ex:"child001"
>
> ex:child001 ex:widgetName ex:"over-ridden widget".
>
> Can I craft a select {?s ?p ?o} that will leave out the link between
> parent and child, yet return the over-ridden values of the predicates?
> i.e., that will return:
> ex:parent001 rdf:type ex:"Parent"
> ex:widgetName ex:"over-ridden widget".
You can do this with COALESCE in SPARQL 1.1:
SELECT ?id COALESCE(?tChild, ?t) AS ?type) ?type (COALESCE(?nChild, ?n) AS ?name)
WHERE {
?id rdf:type ?t .
OPTIONAL {
?id ex:hasChild _:c1 .
_:c1 rdf:type ?tChild .
}
?id ex:widgetName ?n .
OPTIONAL {
?id ex:hasChild _:c2 .
_:c2 ex:widgetName ?nChild .
}
}
This is binding the immediate values, and also binding the child values (where they exist) then returning the child values in preference to the parent ones.
You will get:
?id ?type ?name
ex:parent001 ex:Parent "test"
ex:child001 ex:Parent "over-ridden widget"
The OPTIONAL bits can be written more elegantly like OPTIONAL { ?id ex:hasChild/rdf:type ?tChild } if the store supports property paths.
- Steve
--
Steve Harris, CTO
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Received on Friday, 3 August 2012 09:24:52 UTC