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(a)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.