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.nicolas@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 <thadguidry@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/

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