Lucas_Werkmeister_WMDE added a comment.
That’s true, but Virtuoso also allows this, at least in the DBpedia endpoint <https://dbpedia.org/sparql?query=SELECT+%3Fa+(GROUP_CONCAT(%3Fb%3B+separator+%3D+%22%2C+%22)+AS+%3Fb)+WHERE+{%0D%0A++VALUES+(%3Fa+%3Fb)+{%0D%0A++++(1+%221%22)%0D%0A++++(1+%222%22)%0D%0A++++(2+%223%22)%0D%0A++}%0D%0A}%0D%0AGROUP+BY+%3Fa>. TASK DETAIL https://phabricator.wikimedia.org/T235540 EMAIL PREFERENCES https://phabricator.wikimedia.org/settings/panel/emailpreferences/ To: Lucas_Werkmeister_WMDE Cc: Lea_Lacroix_WMDE, Tagishsimon, Envlh, Mathew.onipe, Smalyshev, Lucas_Werkmeister_WMDE, Igorkim78, Aklapper, Evilricepuddin, darthmon_wmde, DannyS712, Nandana, Lahi, Gq86, GoranSMilovanovic, QZanden, EBjune, merbst, LawExplorer, _jensen, rosalieper, Jonas, Xmlizer, jkroll, Wikidata-bugs, Jdouglas, aude, Tobias1984, Manybubbles, Mbch331
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