FWIW, puoi trovare le query esistenti usando Data Explorer.
Ad esempio: SELECT * from badges WHERE query LIKE '%10 days%'
Ecco quella per 365 giorni:
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
e per 10 giorni:
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
quindi immagino che questa possa funzionare per 30:
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
… che ho verificato funziona e sembra restituire risultati apparentemente sensati, ma non ho controllato approfonditamente i risultati 