5个必考的大厂SQL面试题（复杂的sql语句面试题）

5个必考的大厂SQL面试题（复杂的sql语句面试题）

学Python的同学，SQL也一定要学习，SQL几乎是每个数据岗的必备题目，下面分享几个常见的大厂SQL习题。

（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

（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。喜欢本文记得收藏、关注、点赞。

(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 num from (select user_id,date_sub(log_in_date, rank) dts f rom (select user_id,log_in_date, row_number() over(partitioned by user_id order by log_in_date ) as rank from user_log )t )a group by dts）bwhere num = 7

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

select distinct usr_idfrom ( 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) cgroup by diff,usr_idhaving 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。

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.deptnowhere 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）

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_tfrom 登录信息 as a left join 登录信息 as bon 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 dgroup by d.a_t;

(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 lg2from table) awhere (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_scoreFROM basketball_game_score_detailORDER 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_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_score FROM basketball_game_score_detail ORDER 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_gapfrom ( 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_gap from ( 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_score FROM basketball_game_score_detail ORDER BY SCORE_time ) as x ) as y) as zwhere z.score_gap*last_score_gap<=0and a_sum_score2<>b_sum_score2

​

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