The original code predated SPARQL, so I have to change it anyway. The example I gave is small enough for SPARQL, but others will not be.
On Thu, Jun 1, 2017 at 4:11 PM Daniel Kinzler daniel.kinzler@wikimedia.de wrote:
Am 01.06.2017 um 16:59 schrieb Magnus Manske:
As an example from my BEACON tool, I want all properties that have a
formatter
property, with English label. That SQL is now:
SELECT DISTINCT page_title,term_text FROM pagelinks,page,wb_terms WHERE page_namespace=120 AND substr(page_title,2)=term_entity_id and term_entity_type='property' and term_language='en' and term_type='label'
and
pl_from=page_id and pl_title='P1630' and pl_namespace=120 and pl_from_namespace=120 ORDER BY term_text
Note the "substr". My first attempt was
"page_title=concat('Q',term_entity_id)",
but that took forever.
If we indeed get a full entity ID=page title column for wb_terms, and for wb_items_per_site etc., that would at least fix the on-the-fly
compute. I
shall thus wait with code updates until I get the full story, and not
just
piece-by-piece...
There is currently no plan to put the full ID into wb_items_per_site or wb_property_info, because these tables are bound to a specific entity type. Whether we want to do this would be a whole new discussion.
For what you are doing there, it's probably a lot easier to use the query service. SPARQL:
SELECT DISTINCT ?property ?propertyLabel WHERE { ?property a wikibase:Property . ?property wdt:P1630 ?format . SERVICE wikibase:label { bd:serviceParam wikibase:language "en" } }
-- Daniel Kinzler Principal Platform Engineer
Wikimedia Deutschland Gesellschaft zur Förderung Freien Wissens e.V.