[Niu Ke's questions -sql big factory interview real questions] no2 User growth scenario (a certain degree of information flow)

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Preface

SQL Everyone has to use , But output is not measured by SQL In itself , You need to use this tool , To create other value .

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SQL162 2021 year 11 The average time of browsing articles every day in the month

Create table statement

DROP TABLE IF EXISTS tb_user_log;
CREATE TABLE tb_user_log (
    id INT PRIMARY KEY AUTO_INCREMENT COMMENT ' Self increasing ID',
    uid INT NOT NULL COMMENT ' user ID',
    artical_id INT NOT NULL COMMENT ' video ID',
    in_time datetime COMMENT ' Entry time ',
    out_time datetime COMMENT ' Time to leave ',
    sign_in TINYINT DEFAULT 0 COMMENT ' Check in or not '
) 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);

demand

 Scenario logic description ：artical_id- article ID Represents the of articles viewed by users ID,
artical_id- article ID by 0 Indicates that the user is on the non article content page （ such as App List page in 、 Activity page, etc ）.
 problem ： Statistics 2021 year 11 The average time of browsing articles every day in the month （ Number of seconds ）,
 The result is reserved 1 Decimal place , And sort from short to long on time .

answer

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 The maximum number of people reading each article at the same time

Create table statement

DROP TABLE IF EXISTS tb_user_log;
CREATE TABLE tb_user_log (
    id INT PRIMARY KEY AUTO_INCREMENT COMMENT ' Self increasing ID',
    uid INT NOT NULL COMMENT ' user ID',
    artical_id INT NOT NULL COMMENT ' video ID',
    in_time datetime COMMENT ' Entry time ',
    out_time datetime COMMENT ' Time to leave ',
    sign_in TINYINT DEFAULT 0 COMMENT ' Check in or not '
) 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);

demand

 Scenario logic description ：artical_id- article ID Represents the of articles viewed by users ID,
artical_id- article ID by 0 Indicates that the user is on the non article content page （ such as App List page in 、 Activity page, etc ）.
 problem ： Count the maximum number of people reading each article at the same time , If there are both entry and departure at the same time ,
 First record the increase in the number of users, and then record the decrease , The results are in descending order of the largest number of people .

answer

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 year 11 The next day retention rate of new users every day in the month

Create table statement

DROP TABLE IF EXISTS tb_user_log;
CREATE TABLE tb_user_log (
    id INT PRIMARY KEY AUTO_INCREMENT COMMENT ' Self increasing ID',
    uid INT NOT NULL COMMENT ' user ID',
    artical_id INT NOT NULL COMMENT ' video ID',
    in_time datetime COMMENT ' Entry time ',
    out_time datetime COMMENT ' Time to leave ',
    sign_in TINYINT DEFAULT 0 COMMENT ' Check in or not '
) 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);

demand

 problem ： Statistics 2021 year 11 The next day retention rate of new users every day in the month （ Retain 2 Decimal place ）
 notes ：
 The retention rate of the next day is the proportion of the number of users newly added on the same day and active on the next day .
 If in_time- Entry time and out_time- It's time to leave ,
 It is recorded that the user has been active in two days , The results are in ascending order of date .

answer

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 Count the user rating results of active interval

Create table statement

DROP TABLE IF EXISTS tb_user_log;
CREATE TABLE tb_user_log (
    id INT PRIMARY KEY AUTO_INCREMENT COMMENT ' Self increasing ID',
    uid INT NOT NULL COMMENT ' user ID',
    artical_id INT NOT NULL COMMENT ' video ID',
    in_time datetime COMMENT ' Entry time ',
    out_time datetime COMMENT ' Time to leave ',
    sign_in TINYINT DEFAULT 0 COMMENT ' Check in or not '
) 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);

demand

 problem ： Count the active interval and rank users , Proportion of users at each active level , Two decimal places are reserved for the result , And in descending order of proportion .
 notes ：
 The user level standard is simplified to ： Loyal users ( near 7 Days active and not new users )、 New users ( near 7 Days new )、
 Sleeping users ( near 7 Not active for days, but active earlier )、 Lost users ( near 30 Not active for days, but active earlier ).
 Suppose today is the maximum of all dates in the data .
 near 7 Day means including the day T Close to 7 God , Closed interval [T-6, T].

answer

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 ' Loyal users '
 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 ' New users '
 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 ' Sleeping users '
 else ' Lost users ' end ) user_grade
from tb_user_log
group by uid
) f1
group by user_grade
order by q desc

SQL166 Number of daily activities per day and proportion of new users

Create table statement

DROP TABLE IF EXISTS tb_user_log;
CREATE TABLE tb_user_log (
    id INT PRIMARY KEY AUTO_INCREMENT COMMENT ' Self increasing ID',
    uid INT NOT NULL COMMENT ' user ID',
    artical_id INT NOT NULL COMMENT ' video ID',
    in_time datetime COMMENT ' Entry time ',
    out_time datetime COMMENT ' Time to leave ',
    sign_in TINYINT DEFAULT 0 COMMENT ' Check in or not '
) 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);

demand

 problem ： Count the number of daily activities and the proportion of new users 
 notes ：
 Proportion of new users = Number of new users of the day ÷ Number of active users of the day （ The number of days ）.
 If in_time- Entry time and out_time- It's time to leave , It is recorded that the user has been active in two days .
 The proportion of new users is reserved 2 Decimal place , The results are sorted in ascending order by date .

answer

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 Continuous check-in and receive gold coins

Create table statement

DROP TABLE IF EXISTS tb_user_log;
CREATE TABLE tb_user_log (
    id INT PRIMARY KEY AUTO_INCREMENT COMMENT ' Self increasing ID',
    uid INT NOT NULL COMMENT ' user ID',
    artical_id INT NOT NULL COMMENT ' video ID',
    in_time datetime COMMENT ' Entry time ',
    out_time datetime COMMENT ' Time to leave ',
    sign_in TINYINT DEFAULT 0 COMMENT ' Check in or not '
) 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);

demand

 Scenario logic description ：
artical_id- article ID Represents the of articles viewed by users ID,
 A special case artical_id- article ID by 0 Indicates that the user is on the non article content page （ such as App List page in 、 Activity page, etc ）.
 Be careful ： Only artical_id by 0 when sign_in Value is valid .
 from 2021 year 7 month 7 Japan 0 PM , Users can sign in every day to receive 1 Gold coin , And you can start accumulating check-in days ,
 The second consecutive check-in 3、7 You can receive extra... Every day 2、6 Gold coin .
 Every successive check-in 7 Days to re accumulate check-in days （ That is, reset the check-in days ：
 For the first 8 The day of check-in is recorded as the first day of a new round of check-in , led 1 Gold coin ）
 problem ： Calculate each user 2021 year 7 The number of gold coins obtained each month since the month （ The event ends at 10 The end of the month ,
11 month 1 No more gold coins will be obtained for check-in starting on the th ）. Results by month 、ID Ascending sort .
 notes ： If the check-in record in_time- Entry time and out_time- It's time to leave ,
 Also only recorded as in_time Check in on the corresponding date .

answer

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