FWIW, du kannst die bestehenden Abfragen im Data Explorer finden.
z. B.: SELECT * from badges WHERE query LIKE '%10 days%'
Hier ist die für 365 Tage:
WITH consecutive_visits AS ( SELECT user_id , visited_at , visited_at - (DENSE_RANK() OVER (PARTITION BY user_id ORDER BY visited_at))::int s FROM user_visits ), visits AS ( SELECT user_id , MIN(visited_at) "start" , DENSE_RANK() OVER (PARTITION BY user_id ORDER BY s) "rank" FROM consecutive_visits GROUP BY user_id, s HAVING COUNT(*) >= 365 ) SELECT user_id , "start" + interval '365 days' "granted_at" FROM visits WHERE "rank" = 1
und für 10 Tage:
WITH consecutive_visits AS ( SELECT user_id , visited_at , visited_at - (DENSE_RANK() OVER (PARTITION BY user_id ORDER BY visited_at))::int s FROM user_visits ), visits AS ( SELECT user_id , MIN(visited_at) "start" , DENSE_RANK() OVER (PARTITION BY user_id ORDER BY s) "rank" FROM consecutive_visits GROUP BY user_id, s HAVING COUNT(*) >= 10 ) SELECT user_id , "start" + interval '10 days' "granted_at" FROM visits WHERE "rank" = 1
ich vermute also, dass dies für 30 Tage funktionieren könnte:
WITH consecutive_visits AS ( SELECT user_id , visited_at , visited_at - (DENSE_RANK() OVER (PARTITION BY user_id ORDER BY visited_at))::int s FROM user_visits ), visits AS ( SELECT user_id , MIN(visited_at) "start" , DENSE_RANK() OVER (PARTITION BY user_id ORDER BY s) "rank" FROM consecutive_visits GROUP BY user_id, s HAVING COUNT(*) >= 30 ) SELECT user_id , "start" + interval '30 days' "granted_at" FROM visits WHERE "rank" = 1
… was ich getestet habe und das anscheinend sinnvolle Ergebnisse liefert, aber ich habe die Ergebnisse noch nicht gründlich überprüft 