Yes that is indeed my experience as well.
You can process the results but then in a SELECT within a SELECT query.
This works:
SELECT * WHERE {
{SELECT ?item ?itemLabel WHERE {
?item wdt:P31 wd:Q2085381.
SERVICE wikibase:label { bd:serviceParam wikibase:language
"[AUTO_LANGUAGE],en". }
}}
FILTER(CONTAINS(LCASE(?itemLabel), "simon"))
}
Note the omission of the second FILTER. I suspect that the results of the
service return labels without language tags.
On Fri, May 31, 2019 at 8:12 AM Nicolas VIGNERON <vigneron.nico...@gmail.com>
wrote:
> Hi Thad,
>
> Did it ever works?
> It was my understanding that if you want to manipulate the label (or the
> description, or the alias), you need to explicitly call it and that the
> SERVICE was for display only. At least, this is with this assumption that I
> always wrote my query (or explained during SPARQL workshops) :/
> Anyway, this query works :
>
> SELECT ?item ?itemLabel WHERE {
> ?item wdt:P31 wd:Q2085381 ; rdfs:label ?itemLabel .
> FILTER(CONTAINS(LCASE(?itemLabel), "simon"))
> FILTER (LANG(?itemLabel)="en")
> }
>
> Cheers,
> ~nicolas
>
> Le ven. 31 mai 2019 à 03:05, Thad Guidry <thadgui...@gmail.com> a écrit :
>
>> My Query:
>>
>> SELECT ?item ?itemLabel WHERE {
>> ?item wdt:P31 wd:Q2085381.
>>
>> SERVICE wikibase:label { bd:serviceParam wikibase:language
>> "[AUTO_LANGUAGE],en". }
>> # FILTER(CONTAINS(LCASE(?itemLabel), "simon"))
>> # FILTER (LANG(?itemLabel)="en")
>> }
>>
>> and if I enable any of the FILTER lines, it returns 0 results.
>> What changed / Why ?
>>
>> Thad
>> https://www.linkedin.com/in/thadguidry/
>> _______________________________________________
>> Wikidata mailing list
>> Wikidata@lists.wikimedia.org
>> https://lists.wikimedia.org/mailman/listinfo/wikidata
>>
> _______________________________________________
> Wikidata mailing list
> Wikidata@lists.wikimedia.org
> https://lists.wikimedia.org/mailman/listinfo/wikidata
>
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