Docs please? Can someone point me to the Wikidata docs on this? I'd imagine it's somewhere buried under subpages from https://www.wikidata.org/wiki/Wikidata:SPARQL_query_service/Wikidata_Query_Help ???
Thad https://www.linkedin.com/in/thadguidry/ https://calendly.com/thadguidry/ On Wed, Jan 27, 2021 at 9:04 AM Lucas Werkmeister <[email protected]> wrote: > If you look through the code examples the query service UI generates for > you, you can see that several of them (including the PHP one and the two > JavaScript ones) don’t use any special SPARQL libraries. You can also see > the “URL” option, which is just the URL you’re looking for. The returned > format depends on the Accept request header (or “format” query parameter), > so if you add a request header “Accept: text/csv”, you’ll get CSV back. > > Cheers, > Lucas > On 27.01.21 15:49, Yaron Koren wrote: > > Hi, > > After you run a SPARQL query in the Wikidata Query Service ( > https://query.wikidata.org), the interface provides a lot of options - > you can download the results in formats like CSV and JSON, and it even > provides code to let you run that query directly in a variety of different > computer languages. What I don't know, though, is whether there is some URL > that, when accessed, will directly retrieve the results, in some format > like CSV or JSON. The idea is that some code - in any language - can get > the results without needing any SPARQL-related libraries or special > handling, by just going to the "CSV URL" (for instance) and retrieving the > results. Does such a thing exist? > > -Yaron > > -- > WikiWorks · MediaWiki Consulting · http://wikiworks.com > > _______________________________________________ > Wikidata mailing > [email protected]https://lists.wikimedia.org/mailman/listinfo/wikidata > > _______________________________________________ > Wikidata mailing list > [email protected] > https://lists.wikimedia.org/mailman/listinfo/wikidata >
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