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 ?
Hello,
I also got a similar problem, the following does not return the label:
SELECT ?res ?resLabel WHERE { SELECT ?res ?resLabel WHERE { ?res wdt:P31 wd:Q5 . SERVICE wikibase:label { bd:serviceParam wikibase:language "en". } } LIMIT 1 }
Regards, Fariz
On Fri, May 31, 2019 at 8:05 AM Thad Guidry thadguidry@gmail.com wrote:
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
Hi, I got 8309 results.
Regards
ZI Jony
On Thu, May 30, 2019 at 8:00 PM Fariz Darari fadirra@gmail.com wrote:
Hello,
I also got a similar problem, the following does not return the label:
SELECT ?res ?resLabel WHERE { SELECT ?res ?resLabel WHERE { ?res wdt:P31 wd:Q5 . SERVICE wikibase:label { bd:serviceParam wikibase:language "en". } } LIMIT 1 }
Regards, Fariz
On Fri, May 31, 2019 at 8:05 AM Thad Guidry thadguidry@gmail.com wrote:
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
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/ _______________________________________________ Wikidata mailing list Wikidata@lists.wikimedia.org https://lists.wikimedia.org/mailman/listinfo/wikidata
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/ _______________________________________________ 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
Thanks Andra. Is there no way to proactively without using a 2nd outer query to constrain the query within the WHERE clause? To help with faster query response? In other words, hitting some index so that only publishers containing "simon" are returned.
I'm having to relearn some of this after being away from WDQS for some time.
Thad https://www.linkedin.com/in/thadguidry/
On Fri, May 31, 2019 at 3:56 AM Andra Waagmeester andra@micel.io wrote:
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/ _______________________________________________ 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
Wikidata mailing list Wikidata@lists.wikimedia.org https://lists.wikimedia.org/mailman/listinfo/wikidata
Found my answer !!! had to go back to the SPARQL Tutorial page and re-read the section on FILTER https://www.wikidata.org/wiki/Wikidata:SPARQL_tutorial#FILTER ...specifically...
The label service is very useful if you just want to display the label of a
variable. *But if you want to do stuff with the label* ...
yes, I do want to have an expression about a label..., in fact, 95% of the time that's what I want to do...so why doesn't the darn label service help me do that better?
The reason why this doesn’t work is that the label service adds its
variables very late during query evaluation; at the point where we try to filter on ?humanLabel, the label service hasn’t created that variable yet. Fortunately, the label service isn’t the only way to get an item’s label. Labels are also stored as regular triples, using the predicate rdfs:label. Of course, this means all labels, not just English ones; if we only want English labels, we’ll have to filter on the language of the label:
AH ! label service does things AFTER the query returns results.
So this works, and is how you actually handle label filtering *without* using the label service, and instead getting the label stored as a triple (and it's twice as fast as well)...
SELECT ?publisher ?label WHERE { ?publisher wdt:P31 wd:Q2085381; rdfs:label ?label. FILTER(LANG(?label) = "[AUTO_LANGUAGE]"). FILTER CONTAINS(LCASE(?label), "simon"). }
For this query, you can also use the regular Wikibase search, either skipping the query service completely –
https://www.wikidata.org/wiki/Special:Search/haswbstatement:P31=Q2085381 inlabel:simon@en https://www.wikidata.org/wiki/Special:Search/haswbstatement:P31=Q2085381 inlabel:simon@en https://www.wikidata.org/w/api.php?action=query&list=search&srsearch...
– or combining the query service with the search API via MWAPI https://www.mediawiki.org/wiki/Wikidata_Query_Service/User_Manual/MWAPI:
SELECT ?publisher ?publisherLabel WHERE { SERVICE wikibase:mwapi { bd:serviceParam wikibase:api "Search"; wikibase:endpoint "www.wikidata.org"; mwapi:srsearch "haswbstatement:P31=Q2085381 inlabel:simon@[AUTO_LANGUAGE]". ?publisher wikibase:apiOutputItem mwapi:title. } SERVICE wikibase:label { bd:serviceParam wikibase:language "[AUTO_LANGUAGE],en". } }
On 01.06.19 11:51, Thad Guidry wrote:
Found my answer !!! had to go back to the SPARQL Tutorial page and re-read the section on FILTER https://www.wikidata.org/wiki/Wikidata:SPARQL_tutorial#FILTER...specifically...
The label service is very useful if you just want to display the label of a variable. *But if you want to do stuff with the label** *...yes, I do want to have an expression about a label..., in fact, 95% of the time that's what I want to do...so why doesn't the darn label service help me do that better?
The reason why this doesn’t work is that the label service adds its variables very late during query evaluation; at the point where we try to filter on |?humanLabel|, the label service hasn’t created that variable yet. Fortunately, the label service isn’t the only way to get an item’s label. Labels are also stored as regular triples, using the predicate |rdfs:label|. Of course, this means all labels, not just English ones; if we only want English labels, we’ll have to filter on the language of the label:AH ! label service does things AFTER the query returns results.
So this works, and is how you actually handle label filtering *without* using the label service, and instead getting the label stored as a triple (and it's twice as fast as well)...
SELECT ?publisher ?label WHERE { ?publisher wdt:P31 wd:Q2085381; rdfs:label ?label. FILTER(LANG(?label) = "[AUTO_LANGUAGE]"). FILTER CONTAINS(LCASE(?label), "simon"). }
Thad https://www.linkedin.com/in/thadguidry/
Wikidata mailing list Wikidata@lists.wikimedia.org https://lists.wikimedia.org/mailman/listinfo/wikidata