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【牛客刷题-SQL大厂面试真题】NO2.用户增长场景（某度信息流）
2022-07-01 12:42:00 【IT邦德】
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文章目录
前言
SQL每个人都要用，但是用来衡量产出的并不是SQL本身，你需要用这个工具，去创造其它的价值。
SQL162 2021年11月每天的人均浏览文章时长
建表语句
DROP TABLE IF EXISTS tb_user_log;
CREATE TABLE tb_user_log (
id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
uid INT NOT NULL COMMENT '用户ID',
artical_id INT NOT NULL COMMENT '视频ID',
in_time datetime COMMENT '进入时间',
out_time datetime COMMENT '离开时间',
sign_in TINYINT DEFAULT 0 COMMENT '是否签到'
) CHARACTER SET utf8 COLLATE utf8_bin;
INSERT INTO tb_user_log(uid, artical_id, in_time, out_time, sign_in) VALUES
(101, 9001, '2021-11-01 10:00:00', '2021-11-01 10:00:31', 0),
(102, 9001, '2021-11-01 10:00:00', '2021-11-01 10:00:24', 0),
(102, 9002, '2021-11-01 11:00:00', '2021-11-01 11:00:11', 0),
(101, 9001, '2021-11-02 10:00:00', '2021-11-02 10:00:50', 0),
(102, 9002, '2021-11-02 11:00:01', '2021-11-02 11:00:24', 0);
需求
场景逻辑说明：artical_id-文章ID代表用户浏览的文章的ID，
artical_id-文章ID为0表示用户在非文章内容页（比如App内的列表页、活动页等）。
问题：统计2021年11月每天的人均浏览文章时长（秒数），
结果保留1位小数，并按时长由短到长排序。
答案
select date_format(in_time,"%Y-%m-%d")dt,
round (sum(timestampdiff(second,in_time,out_time))
/count(distinct uid),1) avg_view_len_sec
from tb_user_log
where month(in_time)= 11 and artical_id !=0
group by dt
order by avg_view_len_sec
SQL163 每篇文章同一时刻最大在看人数
建表语句
DROP TABLE IF EXISTS tb_user_log;
CREATE TABLE tb_user_log (
id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
uid INT NOT NULL COMMENT '用户ID',
artical_id INT NOT NULL COMMENT '视频ID',
in_time datetime COMMENT '进入时间',
out_time datetime COMMENT '离开时间',
sign_in TINYINT DEFAULT 0 COMMENT '是否签到'
) CHARACTER SET utf8 COLLATE utf8_bin;
INSERT INTO tb_user_log(uid, artical_id, in_time, out_time, sign_in) VALUES
(101, 9001, '2021-11-01 10:00:00', '2021-11-01 10:00:11', 0),
(102, 9001, '2021-11-01 10:00:09', '2021-11-01 10:00:38', 0),
(103, 9001, '2021-11-01 10:00:28', '2021-11-01 10:00:58', 0),
(104, 9002, '2021-11-01 11:00:45', '2021-11-01 11:01:11', 0),
(105, 9001, '2021-11-01 10:00:51', '2021-11-01 10:00:59', 0),
(106, 9002, '2021-11-01 11:00:55', '2021-11-01 11:01:24', 0),
(107, 9001, '2021-11-01 10:00:01', '2021-11-01 10:01:50', 0);
需求
场景逻辑说明：artical_id-文章ID代表用户浏览的文章的ID，
artical_id-文章ID为0表示用户在非文章内容页（比如App内的列表页、活动页等）。
问题：统计每篇文章同一时刻最大在看人数，如果同一时刻有进入也有离开时，
先记录用户数增加再记录减少，结果按最大人数降序。
答案
SELECT artical_id, max(t1) as t2
FROM(
SELECT artical_id,t,
sum(m)over(partition by artical_id order by t,m desc) as t1
from(
select artical_id,in_time as t,1 as m
from tb_user_log
where artical_id<>0
UNION ALL
select artical_id,out_time as t,-1 as m
from tb_user_log
where artical_id<>0
)a
)b
group by artical_id
order by t2 desc
SQL164 2021年11月每天新用户的次日留存率
建表语句
DROP TABLE IF EXISTS tb_user_log;
CREATE TABLE tb_user_log (
id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
uid INT NOT NULL COMMENT '用户ID',
artical_id INT NOT NULL COMMENT '视频ID',
in_time datetime COMMENT '进入时间',
out_time datetime COMMENT '离开时间',
sign_in TINYINT DEFAULT 0 COMMENT '是否签到'
) CHARACTER SET utf8 COLLATE utf8_bin;
INSERT INTO tb_user_log(uid, artical_id, in_time, out_time, sign_in) VALUES
(101, 0, '2021-11-01 10:00:00', '2021-11-01 10:00:42', 1),
(102, 9001, '2021-11-01 10:00:00', '2021-11-01 10:00:09', 0),
(103, 9001, '2021-11-01 10:00:01', '2021-11-01 10:01:50', 0),
(101, 9002, '2021-11-02 10:00:09', '2021-11-02 10:00:28', 0),
(103, 9002, '2021-11-02 10:00:51', '2021-11-02 10:00:59', 0),
(104, 9001, '2021-11-02 10:00:28', '2021-11-02 10:00:50', 0),
(101, 9003, '2021-11-03 11:00:55', '2021-11-03 11:01:24', 0),
(104, 9003, '2021-11-03 11:00:45', '2021-11-03 11:00:55', 0),
(105, 9003, '2021-11-03 11:00:53', '2021-11-03 11:00:59', 0),
(101, 9002, '2021-11-04 11:00:55', '2021-11-04 11:00:59', 0);
需求
问题：统计2021年11月每天新用户的次日留存率（保留2位小数）
注：
次日留存率为当天新增的用户数中第二天又活跃了的用户数占比。
如果in_time-进入时间和out_time-离开时间跨天了，
在两天里都记为该用户活跃过，结果按日期升序。
答案
select a.first_day, round(count(distinct b.uid)/count(distinct a.uid),2) as rate
from (
select uid, date(min(in_time)) as first_day
from tb_user_log
-- where date_format(in_time, '%Y-%m') = '2021-11'
group by uid
having date_format(min(in_time), '%Y-%m') = '2021-11') a
left join (
select uid, date(in_time) as dt
from tb_user_log
union
select uid, date(out_time) as dt
from tb_user_log) b
on a.uid = b.uid and datediff(b.dt, a.first_day) = 1
group by a.first_day
order by a.first_day
SQL165 统计活跃间隔对用户分级结果
建表语句
DROP TABLE IF EXISTS tb_user_log;
CREATE TABLE tb_user_log (
id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
uid INT NOT NULL COMMENT '用户ID',
artical_id INT NOT NULL COMMENT '视频ID',
in_time datetime COMMENT '进入时间',
out_time datetime COMMENT '离开时间',
sign_in TINYINT DEFAULT 0 COMMENT '是否签到'
) CHARACTER SET utf8 COLLATE utf8_bin;
INSERT INTO tb_user_log(uid, artical_id, in_time, out_time, sign_in) VALUES
(109, 9001, '2021-08-31 10:00:00', '2021-08-31 10:00:09', 0),
(109, 9002, '2021-11-04 11:00:55', '2021-11-04 11:00:59', 0),
(108, 9001, '2021-09-01 10:00:01', '2021-09-01 10:01:50', 0),
(108, 9001, '2021-11-03 10:00:01', '2021-11-03 10:01:50', 0),
(104, 9001, '2021-11-02 10:00:28', '2021-11-02 10:00:50', 0),
(104, 9003, '2021-09-03 11:00:45', '2021-09-03 11:00:55', 0),
(105, 9003, '2021-11-03 11:00:53', '2021-11-03 11:00:59', 0),
(102, 9001, '2021-10-30 10:00:00', '2021-10-30 10:00:09', 0),
(103, 9001, '2021-10-21 10:00:00', '2021-10-21 10:00:09', 0),
(101, 0, '2021-10-01 10:00:00', '2021-10-01 10:00:42', 1);
需求
问题：统计活跃间隔对用户分级后，各活跃等级用户占比，结果保留两位小数，且按占比降序排序。
注：
用户等级标准简化为：忠实用户(近7天活跃过且非新晋用户)、新晋用户(近7天新增)、
沉睡用户(近7天未活跃但更早前活跃过)、流失用户(近30天未活跃但更早前活跃过)。
假设今天就是数据中所有日期的最大值。
近7天表示包含当天T的近7天，即闭区间[T-6, T]。
答案
select user_grade,round(count(uid)
/(select count(distinct uid) from tb_user_log),2) q
from
(
select uid,(case when datediff((select max(in_time) from tb_user_log),max(in_time)) <=6
and datediff((select max(in_time) from tb_user_log),min(in_time)) >6
then '忠实用户'
when datediff((select max(in_time) from tb_user_log),max(in_time)) <=6
and datediff((select max(in_time) from tb_user_log),min(in_time)) <=6
then '新晋用户'
when datediff((select max(in_time) from tb_user_log),max(in_time)) >6
and datediff((select max(in_time) from tb_user_log),min(in_time)) <=29
then '沉睡用户'
else '流失用户' end ) user_grade
from tb_user_log
group by uid
) f1
group by user_grade
order by q desc
SQL166 每天的日活数及新用户占比
建表语句
DROP TABLE IF EXISTS tb_user_log;
CREATE TABLE tb_user_log (
id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
uid INT NOT NULL COMMENT '用户ID',
artical_id INT NOT NULL COMMENT '视频ID',
in_time datetime COMMENT '进入时间',
out_time datetime COMMENT '离开时间',
sign_in TINYINT DEFAULT 0 COMMENT '是否签到'
) CHARACTER SET utf8 COLLATE utf8_bin;
INSERT INTO tb_user_log(uid, artical_id, in_time, out_time, sign_in) VALUES
(101, 9001, '2021-10-31 10:00:00', '2021-10-31 10:00:09', 0),
(102, 9001, '2021-10-31 10:00:00', '2021-10-31 10:00:09', 0),
(101, 0, '2021-11-01 10:00:00', '2021-11-01 10:00:42', 1),
(102, 9001, '2021-11-01 10:00:00', '2021-11-01 10:00:09', 0),
(108, 9001, '2021-11-01 10:00:01', '2021-11-01 10:01:50', 0),
(108, 9001, '2021-11-02 10:00:01', '2021-11-02 10:01:50', 0),
(104, 9001, '2021-11-02 10:00:28', '2021-11-02 10:00:50', 0),
(106, 9001, '2021-11-02 10:00:28', '2021-11-02 10:00:50', 0),
(108, 9001, '2021-11-03 10:00:01', '2021-11-03 10:01:50', 0),
(109, 9002, '2021-11-03 11:00:55', '2021-11-03 11:00:59', 0),
(104, 9003, '2021-11-03 11:00:45', '2021-11-03 11:00:55', 0),
(105, 9003, '2021-11-03 11:00:53', '2021-11-03 11:00:59', 0),
(106, 9003, '2021-11-03 11:00:45', '2021-11-03 11:00:55', 0);
需求
问题：统计每天的日活数及新用户占比
注：
新用户占比=当天的新用户数÷当天活跃用户数（日活数）。
如果in_time-进入时间和out_time-离开时间跨天了，在两天里都记为该用户活跃过。
新用户占比保留2位小数，结果按日期升序排序。
答案
select dt, count(*) as dau, round(sum(new)/count(*), 2) as uv_new_ratio
from (
select uid, dt, case when dt = first_dt then 1 else 0 end as new
from
(select uid, date(in_time) as dt
from tb_user_log
UNION
select uid, date(out_time) as dt
from tb_user_log) t1
left join
(select uid, min(date(in_time)) as first_dt
from tb_user_log
group by uid) t2
using(uid)
) t
group by dt
order by dt
SQL167 连续签到领金币
建表语句
DROP TABLE IF EXISTS tb_user_log;
CREATE TABLE tb_user_log (
id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
uid INT NOT NULL COMMENT '用户ID',
artical_id INT NOT NULL COMMENT '视频ID',
in_time datetime COMMENT '进入时间',
out_time datetime COMMENT '离开时间',
sign_in TINYINT DEFAULT 0 COMMENT '是否签到'
) CHARACTER SET utf8 COLLATE utf8_bin;
INSERT INTO tb_user_log(uid, artical_id, in_time, out_time, sign_in) VALUES
(101, 0, '2021-07-07 10:00:00', '2021-07-07 10:00:09', 1),
(101, 0, '2021-07-08 10:00:00', '2021-07-08 10:00:09', 1),
(101, 0, '2021-07-09 10:00:00', '2021-07-09 10:00:42', 1),
(101, 0, '2021-07-10 10:00:00', '2021-07-10 10:00:09', 1),
(101, 0, '2021-07-11 23:59:55', '2021-07-11 23:59:59', 1),
(101, 0, '2021-07-12 10:00:28', '2021-07-12 10:00:50', 1),
(101, 0, '2021-07-13 10:00:28', '2021-07-13 10:00:50', 1),
(102, 0, '2021-10-01 10:00:28', '2021-10-01 10:00:50', 1),
(102, 0, '2021-10-02 10:00:01', '2021-10-02 10:01:50', 1),
(102, 0, '2021-10-03 11:00:55', '2021-10-03 11:00:59', 1),
(102, 0, '2021-10-04 11:00:45', '2021-10-04 11:00:55', 0),
(102, 0, '2021-10-05 11:00:53', '2021-10-05 11:00:59', 1),
(102, 0, '2021-10-06 11:00:45', '2021-10-06 11:00:55', 1);
需求
场景逻辑说明：
artical_id-文章ID代表用户浏览的文章的ID，
特殊情况artical_id-文章ID为0表示用户在非文章内容页（比如App内的列表页、活动页等）。
注意：只有artical_id为0时sign_in值才有效。
从2021年7月7日0点开始，用户每天签到可以领1金币，并可以开始累积签到天数，
连续签到的第3、7天分别可额外领2、6金币。
每连续签到7天后重新累积签到天数（即重置签到天数：
连续第8天签到时记为新的一轮签到的第一天，领1金币）
问题：计算每个用户2021年7月以来每月获得的金币数（该活动到10月底结束，
11月1日开始的签到不再获得金币）。结果按月份、ID升序排序。
注：如果签到记录的in_time-进入时间和out_time-离开时间跨天了，
也只记作in_time对应的日期签到了。
答案
SELECT uid,DATE_FORMAT(sign_dt,'%Y%m')as month,sum(coin)
FROM
(SELECT uid,sign_dt,TIMESTAMPADD(day,-diff+1,sign_dt)as start_day ,
case (DENSE_RANK() over (PARTITION by uid,TIMESTAMPADD(day,-diff+1,sign_dt) ORDER BY sign_dt))%7
WHEN 3 then 3
WHEN 0 THEN 7
ELSE 1 end as coin
FROM
(SELECT uid ,DATE_FORMAT(in_time,'%Y%m%d')as sign_dt,
DENSE_RANK() over(PARTITION by uid ORDER BY in_time) as diff
FROM tb_user_log
WHERE DATE_FORMAT(in_time,'%Y%m%d') BETWEEN 20210707 and 20211031
AND artical_id =0 AND sign_in =1 )t1 )t2
GROUP BY uid,DATE_FORMAT(sign_dt,'%Y%m')
ORDER BY DATE_FORMAT(sign_dt,'%Y%m') ,uid
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