五大常考SQL面试题

目录

一、找出连续7天登陆，连续30天登陆的用户（小红书笔试，电信云面试），最大连续登陆天数的问题 --窗口函数

 二、求连续点击三次的用户数，而且中间不能有别人的点击

三、计算除去部门最高工资，和最低工资的平均工资（字节跳动面试）--窗口函数     

四、留存的计算，和累计求和的计算 --窗口函数，自联结（pdd面试）

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一、找出连续7天登陆，连续30天登陆的用户（小红书笔试，电信云面试），最大连续登陆天数的问题 --窗口函数

        1、先创建表以及导入数据

create table login(
user_id int comment '用户id',
access_time datetime comment '访问时间',
page_id int comment '页面id',
dt date comment '登陆日期'
);

insert into login values
(1, '2021-06-01 11:13:15', 10, '2021-06-01'),
(1, '2021-06-02 11:13:15', 10, '2021-06-02'),
(1, '2021-06-03 11:13:15', 10, '2021-06-03'),
(1, '2021-06-04 11:13:15', 10, '2021-06-04'),
(1, '2021-06-05 11:13:15', 10, '2021-06-05'),
(1, '2021-06-06 11:13:15', 10, '2021-06-06'),
(1, '2021-06-07 11:13:15', 10, '2021-06-07'),
(2, '2021-06-01 11:13:15', 10, '2021-06-01'),
(2, '2021-06-03 11:13:15', 10, '2021-06-03'),
(2, '2021-06-04 11:13:15', 10, '2021-06-04'),
(2, '2021-06-05 11:13:15', 10, '2021-06-05'),
(3, '2021-06-01 11:13:15', 10, '2021-06-01'),
(3, '2021-06-07 11:13:15', 10, '2021-06-07'),
(3, '2021-06-08 11:13:15', 10, '2021-06-08'),
(3, '2021-06-09 11:13:15', 10, '2021-06-09'),
(3, '2021-06-10 11:13:15', 10, '2021-06-10'),
(3, '2021-06-11 11:13:15', 10, '2021-06-11'),
(3, '2021-06-12 11:13:15', 10, '2021-06-12'),
(3, '2021-06-13 11:13:15', 10, '2021-06-13'),
(4, '2021-06-01 11:13:15', 10, '2021-06-01'),
(4, '2021-06-03 11:13:15', 10, '2021-06-03'),
(4, '2021-06-05 11:13:15', 10, '2021-06-05'),
(4, '2021-06-07 11:13:15', 10, '2021-06-07'),
(4, '2021-06-09 11:13:15', 10, '2021-06-09'),
(4, '2021-06-11 11:13:15', 10, '2021-06-11'),
(5, '2021-06-01 11:13:15', 10, '2021-06-01'),
(5, '2021-06-07 11:13:15', 10, '2021-06-07'),
(5, '2021-06-08 11:13:15', 10, '2021-06-08'),
(5, '2021-06-09 11:13:15', 10, '2021-06-09'),
(5, '2021-06-11 11:13:15', 10, '2021-06-11'),
(5, '2021-06-12 11:13:15', 10, '2021-06-12'),
(5, '2021-06-13 11:13:15', 10, '2021-06-13');

        2、思路

                ①先将用户按照id进行分组，然后再用窗口函数进行排序

SELECT *,ROW_NUMBER() over (PARTITION by user_id order by user_id)FROM login where month(dt) = 6

                ②将日期与排名做差 （如果用户是连续登陆，那么差值日期的结果是一样的）

select *,date_sub(dt,INTERVAL ranking day) diff from (SELECT *,ROW_NUMBER() over (PARTITION by user_id order by dt) ranking FROM login where month(dt) = 6) t

               

                 ③根据用户与时间差分类，统计出现的次数，如果次数在7日以上，则是想要的结果

select user_id ,count(*) from
(select *, date_sub(dt, interval ranking day) diff from
(select *, row_number() over(partition by user_id order by dt) ranking from login where month(dt)=6) as t) as t1
group by user_id, diff having count(*) >= 7;

 二、求连续点击三次的用户数，而且中间不能有别人的点击

        a表记录了点击的流水信息，包括用户id ，和点击时间

row_number() over(order by click_time) as rank_1 得到rank_1为 1 2 3 4 5 6 7

row_number() over(partition by usr_id order by click_time) 得到rank_2 为 1 2 1 3 4 5 6

rank_1- rank2 得到diff 为 0 0 2 1 1 1 1

        这时我们发现只需要对diff进行分组计数大于3个，就是连续点击大于三且中间没有其他人点击的用户

select distinct usr_id
from    
(
   select *, rank_1- rank2  as diff
   from
  (
      select *,
      row_number() over(order by click_time) as  rank_1
      row_number() over(partition by usr_id order by click_time) as rank_2
      from a
   ) b
) c
group by diff,usr_id
having count(diff) >=3

三、计算除去部门最高工资，和最低工资的平均工资（字节跳动面试）--窗口函数     

        emp 表

  id 员工 id ，deptno 部门编号，salary 工资

        核心是使用窗口函数降序和升序分别排一遍就取出了最高和最低。

select a.deptno，avg(a.salary)
from  
 (
 select *, rank() over( partition by deptno order by salary ) as rank_1
 , rank() over( partition by deptno order by salary desc) as rank_2 
 from emp
 )  a 
group by a.deptno
where a.rank_1 >1 and a.rank_2 >1 

四、留存的计算，和累计求和的计算 --窗口函数，自联结（pdd面试）

        手机中的相机是深受大家喜爱的应用之一，下图是某手机厂商数据库中的用户行为信息表中部分数据的截图

        现在该手机厂商想要分析手机中的应用（相机）的活跃情况，需统计如下数据： 

需要获得的数据的格式如下：

select d.a_t,count(distinct case when d.时间间隔=1 then d.用户id     
               else null
               end) as  次日留存数, 
count(distinct case when 时间间隔=1 then d.用户id
               else null
               end) /count(distinct d.用户id) as 次日留存率,
count(distinct case when d.时间间隔=3 then d.用户id     
               else null
               end) as  3日留存数 ,
count(distinct case when 时间间隔=3 then d.用户id
               else null
               end) /count(distinct d.用户id) as 3日留存率,
count(distinct case when d.时间间隔=7 then d.用户id     
               else null
               end) as  7日留存数 ,
count(distinct case when 时间间隔=7 then d.用户id
               else null
               end) /count(distinct d.用户id) as 7日留存率

from
(select *,timestampdiff(day,a_t,b_t) as 时间间隔
from (select a.`用户id`,a.登陆时间 as a_t ,b.登陆时间 as b_t
from 登录信息 as a  
left join 登录信息 as b
on a.`用户id`=b.`用户id`
where a.应用名称= '相机' AND b.应用名称='相机') as c) as d
group by d.a_t;