顺便一提，你可以使用数据探索器找到现有的查询。
例如：SELECT * from badges WHERE query LIKE '%10 days%'
以下是针对 365 天的查询：
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
以及 10 天的查询：
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
所以我猜这个查询应该也能适用于 30 天：
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
……我测试过这个查询可以运行，返回的结果看起来也合理，但我还没有仔细验证结果 