数据分析常见SQL面试笔试题

（1）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
（2）求连续点击三次的用户数，中间不能有别人的点击，最大连续天数的变形问题（腾讯微保面试）–窗口函数
（3）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
（4）留存的计算，和累计求和的计算 --窗口函数，自联结（pdd面试）
（5）AB球队得分流水表，得到连续三次得分的队员名字和每次赶超对手的球员名字,（pdd面试）把这几类题型吃透，再也不怕手撕SQL和笔试了，其中最难的是题
（5）,整个面试的sql基本上都是窗口函数的玩法，搭配case when 也考得比较多。

(1) 找出连续7天登陆，连续30天登陆的用户

select *
from
（select user_id ,count(1) as numfrom(select user_id,date_sub(log_in_date, rank) dtsf rom  (select user_id,log_in_date,   row_number() over(partitioned by user_id order by log_in_date ) as rankfrom user_log)t)agroup by dts
）b
where num = 7

(2)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usr_id a a b a a a aclick_time t1 t2 t3 t4 t5 t6 t7 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 usrid
from
(select *, rank_1- rank2  as difffrom(select *,row_number() over(order by click_time) as  rank_1row_number() over(partition by usr_id order by click_time) as rank_2from a) b
) c
group by diff,usr_id
having count(diff) >=3

(3)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。

elect 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

(4) 留存的计算，和累计求和的计算 --窗口函数，自联结（pdd面试）累计求和问题，使用sum() 窗口函数留存计算绚丽的小海螺：SQL 计算留存率-链家面试题(5)AB球队得分流水表，得到连续三次得分的队员名字和每次赶超对手的球员名字（pdd）在复盘时发现有类似原题，这是我在面试中遇到的最难的题问题：两支篮球队进行了激烈的篮球比赛，比分交替上升。比赛结束后，你有一张两队得分分数的明细表，记录了球队team，球员号码number，球员姓名name, 得分分数score 以及得分时间scoretime(datetime)。现在球队要对比赛中表现突出的球员做出嘉奖，所以请你用sql统计出1)连续三次（及以上）为球队得分的球员名单2)比赛中帮助各自球队反超比分的球员姓名以及对应时间。先建一个类似的表

CREATE TABLE basketball_game_score_detail(team  VARCHAR(40) NOT NULL ,number VARCHAR(100) NOT NULL,score_time datetime NOT NULL,score int NOT NULL,name varchar(100)  NOT NULL
);
insert into  basketball_game_score_detail values('A',1,'2020/8/28 9:01:14',1,'A1');
insert into  basketball_game_score_detail values('A',5,'2020/8/28 9:02:28',1,'A5');
insert into  basketball_game_score_detail values('B',4,'2020/8/28 9:03:42',3,'B4');
insert into  basketball_game_score_detail values('A',4,'2020/8/28 9:04:55',3,'A4');
insert into  basketball_game_score_detail values('B',1,'2020/8/28 9:06:09',3,'B1');
insert into  basketball_game_score_detail values('A',3,'2020/8/28 9:07:23',3,'A3');
insert into  basketball_game_score_detail values('A',4,'2020/8/28 9:08:37',3,'A4');
insert into  basketball_game_score_detail values('B',1,'2020/8/28 9:09:51',2,'B1');
insert into  basketball_game_score_detail values('B',2,'2020/8/28 9:11:05',2,'B2');
insert into  basketball_game_score_detail values('B',4,'2020/8/28 9:12:18',1,'B4');
insert into  basketball_game_score_detail values('A',1,'2020/8/28 9:13:32',2,'A1');
insert into  basketball_game_score_detail values('A',1,'2020/8/28 9:14:46',1,'A1');
insert into  basketball_game_score_detail values('A',4,'2020/8/28 9:16:00',1,'A4');
insert into  basketball_game_score_detail values('B',3,'2020/8/28 9:17:14',3,'B3');
insert into  basketball_game_score_detail values('B',2,'2020/8/28 9:18:28',3,'B2');
insert into  basketball_game_score_detail values('A',2,'2020/8/28 9:19:42',3,'A2');
insert into  basketball_game_score_detail values('A',1,'2020/8/28 9:20:55',1,'A1');
insert into  basketball_game_score_detail values('B',3,'2020/8/28 9:22:09',2,'B3');
insert into  basketball_game_score_detail values('B',3,'2020/8/28 9:23:23',3,'B3');
insert into  basketball_game_score_detail values('A',5,'2020/8/28 9:24:37',2,'A5');
insert into  basketball_game_score_detail values('B',1,'2020/8/28 9:25:51',3,'B1');
insert into  basketball_game_score_detail values('B',2,'2020/8/28 9:27:05',1,'B2');
insert into  basketball_game_score_detail values('A',3,'2020/8/28 9:28:18',1,'A3');
insert into  basketball_game_score_detail values('B',4,'2020/8/28 9:29:32',1,'B4');
insert into  basketball_game_score_detail values('A',1,'2020/8/28 9:30:46',3,'A1');
insert into  basketball_game_score_detail values('B',1,'2020/8/28 9:32:00',1,'B1');
insert into  basketball_game_score_detail values('A',4,'2020/8/28 9:33:14',2,'A4');
insert into  basketball_game_score_detail values('B',1,'2020/8/28 9:34:28',1,'B1');
insert into  basketball_game_score_detail values('B',5,'2020/8/28 9:35:42',2,'B5');
insert into  basketball_game_score_detail values('A',1,'2020/8/28 9:36:55',1,'A1');
insert into  basketball_game_score_detail values('B',1,'2020/8/28 9:38:09',3,'B1');
insert into  basketball_game_score_detail values('A',1,'2020/8/28 9:39:23',3,'A1');
insert into  basketball_game_score_detail values('B',2,'2020/8/28 9:40:37',3,'B2');
insert into  basketball_game_score_detail values('A',3,'2020/8/28 9:41:51',3,'A3');
insert into  basketball_game_score_detail values('A',1,'2020/8/28 9:43:05',2,'A1');
insert into  basketball_game_score_detail values('B',3,'2020/8/28 9:44:18',3,'B3');
insert into  basketball_game_score_detail values('A',5,'2020/8/28 9:45:32',2,'A5');
insert into  basketball_game_score_detail values('B',5,'2020/8/28 9:46:46',3,'B5');

这里我使用了lead和lag来取每个组的前几个值，这个和最大联系天数不太一样，但也可以用类似思路去解，但是使用lead和lag做起来更容易理解

select distinct a.name ,a.team from
(
select *,lead(name,1) over(partition by team order by score_time) as ld1
,lead(name,2) over(partition by team order by score_time) as ld2
,lag(name,1) over(partition by team order by score_time) as lg1
,lag(name,2) over(partition by team order by score_time) as lg2
from table
) a
where (a.name =a.ld1 and a.name =a.ld2)
or (a.name =a.ld1 and a.name =a.lg1)
or (a.name=a.lg1 and a.name=a.lg2)

第二小问面试时没完全做出来，说了下思路，现在想了想当时的思路还是有问题，而且这个题也并不难，核心还是记录每个时刻的累计得分表

SELECT TEAM,number,name,score_time,score,case when team='A' then score else 0 end as A_score
,case when team='B' then score else 0 end B_score
FROM basketball_game_score_detail
ORDER BY SCORE_time

如下得到每个时刻的累计得分表

select team,number,name,score_time,A_score,b_score
,sum(A_score)over(order by score_time) as  a_sum_score2
,sum(b_score)over(order by score_time) as b_sum_score2
from
(SELECT TEAM,number,name,score_time,score,case when team='A' then score else 0 end as A_score,case when team='B' then score else 0 end B_scoreFROM basketball_game_score_detailORDER BY SCORE_time
) as x

计算每个时刻的累计得分差，和上个时间的累计得分差，只要两个的符号相反就是反超时刻。感觉思路还是比较简洁的。

select *,score_gap*last_score_gap
from
(select  *,a_sum_score2-b_sum_score2 as score_gap ,lag(a_sum_score2-b_sum_score2,1)over(order by score_time) as last_score_gapfrom (select team,number,name,score_time,A_score,b_score,sum(A_score)over(order by score_time) as  a_sum_score2,sum(b_score)over(order by score_time) as b_sum_score2from (SELECT TEAM,number,name,score_time,score,case when team='A' then score else 0 end as A_score,case when team='B' then score else 0 end B_scoreFROM basketball_game_score_detailORDER BY SCORE_time) as x) as y
) as z
where z.score_gap*last_score_gap<=0
and a_sum_score2<>b_sum_score2

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