sql常考题

留存率问题
牛客上的原题https://www.nowcoder.com/practice/1fc0e75f07434ef5ba4f1fb2aa83a450?tpId=268&tqId=2285344&ru=/exam/oj&qru=/ta/sql-factory-interview/question-ranking&sourceUrl=%2Fexam%2Foj%3Ftab%3DSQL%25E7%25AF%2587%26topicId%3D268
解答

select a.dt,
       round(count(b.uid)/ count(a.uid),2) as uv_left_rate
from (select uid
      ,min(date(in_time)) dt
      from tb_user_log
      group by uid) as a
left join (select uid , date(in_time) dt
           from tb_user_log
           union
           select uid , date(out_time)
           from tb_user_log) as b 
on a.uid=b.uid
and a.dt=date_sub(b.dt,INTERVAL 1 day)
where date_format(a.dt,"%Y-%m")='2021-11'
group by a.dt

进阶：

/*计算某日新增登录设备的次日、3日、7日、14日、30日、90日留存率*/
SELECT
	log_day '日期',
	count(user_id_d0) '新增数量',
	count(user_id_d1) / count(user_id_d0) '次日留存',
	count(user_id_d3) / count(user_id_d0) '3日留存',
	count(user_id_d7) / count(user_id_d0) '7日留存',
	count(user_id_d14) / count(user_id_d0) '14日留存',
	count(user_id_d30) / count(user_id_d0) '30日留存',
	count(user_id_d90) / count(user_id_d0) '90日留存'
FROM
	(
		SELECT DISTINCT
			log_day,
			a.user_id_d0,
			b.device_id AS user_id_d1,
			c.device_id AS user_id_d3,
			d.device_id AS user_id_d7,
			e.device_id AS user_id_d14,
			f.device_id AS user_id_d30,
			g.device_id AS user_id_d90
		FROM
			(
				SELECT DISTINCT
					Date(event_time) AS log_day,
					device_id AS user_id_d0
				FROM
					role_login_back
				GROUP BY
					device_id
				ORDER BY
					log_day
			) a
		LEFT JOIN role_login_back b ON DATEDIFF(DATE(b.event_time),a.log_day) = 1
		AND a.user_id_d0 = b.device_id
		LEFT JOIN role_login_back c ON DATEDIFF(DATE(c.event_time),a.log_day) = 2
		AND a.user_id_d0 = c.device_id
		LEFT JOIN role_login_back d ON DATEDIFF(DATE(d.event_time),a.log_day) = 6
		AND a.user_id_d0 = d.device_id
		LEFT JOIN role_login_back e ON DATEDIFF(DATE(e.event_time),a.log_day) = 13
		AND a.user_id_d0 = e.device_id
		LEFT JOIN role_login_back f ON DATEDIFF(DATE(f.event_time),a.log_day) = 29
		AND a.user_id_d0 = f.device_id
		LEFT JOIN role_login_back g ON DATEDIFF(DATE(g.event_time),a.log_day) = 89
		AND a.user_id_d0 = g.device_id
	) AS temp
GROUP BY
log_day

原文：https://zhuanlan.zhihu.com/p/143494489