用户留存率SQL题

Posted 2022-08-01 wzy0623

原文：牛客每个人最近的登录日期(五)_牛客题霸_牛客网

-- 方法一
SELECT t0.date, IFNULL(ROUND(COUNT(DISTINCT t2.user_id) / (COUNT(t1.user_id)),3),0)
  FROM (SELECT date FROM login GROUP BY date) t0 
  LEFT JOIN (SELECT user_id, MIN(date) AS date FROM login GROUP BY user_id) t1 ON t0.date = t1.date 
  LEFT JOIN login AS t2 ON t1.user_id = t2.user_id AND DATEDIFF(t2.date, t1.date) = 1
 GROUP BY t0.date;

-- 方法二
SELECT date, IFNULL(ROUND((SUM(CASE WHEN (user_id , date) IN (SELECT user_id, DATE_ADD(date, INTERVAL - 1 DAY) FROM login) AND 
                                         (user_id , date) IN (SELECT user_id, MIN(date) FROM login GROUP BY user_id)
                                    THEN 1
                                    ELSE 0
                                END)) / 
                          (SUM(CASE WHEN (user_id , date) IN (SELECT user_id, MIN(date) FROM login GROUP BY user_id)
                                    THEN 1
                                    ELSE 0
                                END)),
                         3),0) AS p
  FROM login
 GROUP BY date
 ORDER BY date;

（注：方法一的3表关联对大表来说必须加索引，方法二的子查询方式不推荐

create index idx_date on login(date);
create index idx_user_id_date on login(user_id,date);

）