一个表table: c_id u_id score；用SQL计算每个班级top5学生的平均分（腾讯）
select class_id,avg(score) as score_avg
from
(select *,row_number() over(partition by class_id order by score desc) as score_rank
from table
) t1
where t1.score_rank<=5
gropu by t1.class_id
计算连续登陆3天的用户？营业额连续增长的店铺？（腾讯、零食有鸣）
--原始数据
u_1 2024-08-01
u_1 2024-08-02
u_1 2024-08-03
u_1 2024-08-05
u_1 2024-08-06u_2 2024-08-01
u_2 2024-08-03
--lag() over()函数+dadediflag(date,1) diff_1 lag(date,2). diff_2
u_1 2024-08-01 null null null null
u_1 2024-08-02 2024-08-01 1 null null
u_1 2024-08-03 2024-08-02 1 2024-08-01 1
u_1 2024-08-05 2024-08-03 2 2024-08-02 1
u_1 2024-08-06 2024-08-05 1 2024-08-03 2u_2 2024-08-01 null 1 null null
u_2 2024-08-03 2024-08-01 2 null nullselect distinct user_id
from table
where lag(date,1) is not nulland diff_1=1and diff_2=1
腾讯SQL手写笔试题目
1.
先会有登陆事件、再有对局事件
login_log表作为left join的主表一个玩家登陆之后，可能会对应多次对局
对局时间大于最近一次登陆时间：每一个对局时间，对应多个登陆时间，找到最晚的那一个就可以了select t1.uid,t2.battle_time,max(login_time) as login_time
from login_log t1
left join battle_log t2
on t1.uid=t2.uid
and t1.login_time<t2.battle_time
group by t1.uid,t2.battle_time2.
首先进行split切分操作
其次进行行转列操作，将数据打散
最后按照兴趣聚合select hobby_new_1,count(name) as user_cnt
from
(
select t1.name,t2.hobby_new_1
from
(select name,split(hobby,'+') as hobby_new
from hobby_detail
) t1
lateral view explode(t1.hobby_new) hobby_new_1 as t2
) t
group by hobby_new_1
腾讯SQL手写笔试题目
问题1
引擎：hive/spark
语句：
select stu_id
from scoce_detail
group by stu_id
having min(score) > 60;
思路：在scoce_detail表中，每一个学生最小的成绩只要大于60分，那么其他科目都满足条件，该学生就是目标学生（前提是学生某些科目没有成绩，不做过滤）问题2
引擎：hive/spark
语句：
select t.stu_id
from
(select stu_id,count(if(subject='数学',1,null)) as math_subject_cnt
from score_detail
group by stu_id
having(count(if(subject='数学',1,null)))<1
) t ;
思路：统计每个学生下的 数学科目的条数，如果数学科目条数小于1，代表该学生没有数据成绩
