Does this address your question:
SELECT ?projectGraph ?status ?statusLabel
WHERE {
rdf:nil teamwork:graphsUnderTeamControl ( ?projectGraph ) .
GRAPH ?projectGraph {
?projectGraph a
<http://evn.topbraidlive.org/evnprojects#Taxonomy> .
?projectGraph <http://topbraid.org/metadata#status> ?status .
GRAPH <http://topbraid.org/metadata> {
?status rdfs:label ?statusLabel
}
} .
}
The trick here is to get the display label of the status resource, but
from the graph where it is stored.
Regards,
Holger
On 29/10/2018 7:39 AM, William Ramos wrote:
Hi, I'm relatively new to using SPARQL and the Topbraid composer. I've
been experimenting on the pre-built example taxonomies such as
Geography and the IPTC news codes.
My specific question is about how to properly retrieve the "status"
value of all the taxonomies, for example, on the EDG application,
inside a taxonomy (in this case Geography); under the "Dashboard" tab
we can see the following two fields:
*Description and status. *After using the following query I'm able to
retrieve the "Description" values as-is, however for "status" I get
the URI like "<http://topbraid.org/metadata#InUseStatus>" instead of
the expected value, which in this case would be "In use" exactly as
shown on the EDG view above. Please see below the query and results I get:
*SPARQL Query:*
SELECT ?p ?o ?projectGraph
WHERE {
rdf:nil teamwork:graphsUnderTeamControl ( ?projectGraph ) .
GRAPH ?projectGraph {
?projectGraph a
<http://evn.topbraidlive.org/evnprojects#Taxonomy> .
?projectGraph ?p ?o .
FILTER (?p = <http://topbraid.org/metadata#status> || ?p =
rdfs:comment) .
} .
}
*Results:*
*
*
Thanks.
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