2.用户增长场景（某度信息流）
SQL162 2021年11月每天的人均浏览文章时长
select
date_format(in_time,'%Y-%m-%d') as dt,
round(sum(timestampdiff(second,in_time,out_time))/count(distinct uid),1) as avg_viiew_len_sec
from
tb_user_log
where date_format(in_time,'%Y-%m')='2021-11' and artical_id !=0
group by dt
order by avg_viiew_len_sec
SQL163 每篇文章同一时刻最大在看人数
( 1.’ 1 diff ’ 表示的意思为增加列名为diff 的临时列，值全为1 ；
2. 题目要求在瞬时统计时遵循【先进后出】：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。
因此在ORDER BY层面，在遵循dt升序的同时，还要遵循先+1，再-1的原则，即diff DESC：
SUM(diff) OVER(PARTITION BY artical_id ORDER BY dt, diff DESC) ）
select
artical_id,
max(instant_uv) as max_uv
from(
select artical_id,
sum(diff)over(partition by artical_id order by dt,diff desc) as instant_uv
from (
select artical_id,in_time as dt, 1 diff
from tb_user_log
where artical_id !=0
union all
select artical_id,out_time as dt, -1 diff
from tb_user_log
where artical_id !=0 ) t1
) t2
group by artical_id
order by max_uv desc
SQL164 2021年11月每天新用户的次日留存率
(1.DATE_SUB() 函数指的是从日期减去指定的时间间隔,
interval为取间隔的含义，在这个SQL中，子查询的where条件中含义说白了就是：
当A表时间 = B表时间 - 1天
2.倒数第四行a.dt=date_sub(b.dt,interval 1 day)前面为什么要用and连接，而不用where呢？
on 表示的是两个表连接的依据（本题中表连接的依据是UID相等且日期相差一天），
where表示的是进行筛选，
在本题目中，通过where的子句进行筛选，第二天未活跃的数据都过滤掉了。
)
select a.dt,
round(count(b.uid)/count(a.uid),2) as uv_left_rate
from(
select uid,
min(date(in_time)) as dt
from tb_user_log
group by uid
) as 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
) as b
on a.uid=b.uid
and a.dt=date_sub(b.dt,interval 1 day)
where date_format(a.dt,'%Y%m')='202111'
group by a.dt
order by a.dt
SQL165 统计活跃间隔对用户分级结果
select user_grade,
round(count(uid)/(select count(distinct uid) from tb_user_log ),2) as ratio
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))>=7
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 ) as user_grade
from tb_user_log
group by uid
) as a
group by user_grade
order by ratio desc
SQL166 每天的日活数及新用户占比
select dt,
count(*) as dau,
round(sum(new)/count(*),2)
from(
select dt,uid,
case when dt=first 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
) as a
left join
(
select uid,
min(date(in_time)) as first
from tb_user_log
group by uid
) as b
using(uid)
) as c
group by dt
order by dt
SQL167 连续签到领金币
(思路如下：
1.%7 是什么意思
% 号的意思是取余，%7表示除以7的余数，3除以7 等于 0 余3, 8除以7等于1 余 1;
2. 对上图的补充解释：
本题的难点在于怎么求得连续性，通过a表中的row_number排序，给每个dt一个对应的序号；
然后通过b表中的subdate求得签到日期dt与row_number得出的序列号的差值，如果差值不变，即签到日期与序列同步变化，提升一天，序号增加1，则表示他在连续签到; 提升多天，序号增加1，差值不恒定，则表示他没有连续签到；
通过c表中的row_number给差值分组进行排序，差值不变，即连续签到，则单独分组进行排序；差值变动，则表示没有连续签到，另外分一组再次进行排序。对结果用%取余数；
最后用case when 分情况讨论，余数为三时，加三个积分。余数为0时，表示连续了七天，加七个积分。）
select uid,date_format(dt,'%Y%m') as month
,sum(case when stage_index=3 then 3
when stage_index=0 then 7
else 1 end) as coin
from(
select uid,dt
,(row_number()over(partition by uid,init_date order by dt))%7 as stage_index
from
(
select uid,dt,rn,subdate(dt,rn) as init_date
from (
select uid,date(in_time) as dt
,row_number()over(partition by uid order by date(in_time)) as rn
from tb_user_log
where date(in_time) >='2021-07-07'
and date(in_time)<'2021-11-01'
and artical_id=0 and sign_in=1
) as a
) as b
) as c
group by uid,month
order by month,uid;