本文提及技巧只适用数据统计，OLAP场景，不建议用于业务sql。

分组top问题

知识点：巧用局部变量，实现Hive窗口分析函数功能row_number() over(partition by exam_id)

题目：如下表，记录每个学生的成绩，用sql求出每班前五名的学生id和成绩。

create table student(  id int(11) comment "学生id",  class int(3) comment "班级",  score int(3)  comment "成绩");

解答1：巧用临时变量

划重点：mysql的临时变量赋值顺序是：from --> select，select子句字段按从上到下顺序执行。

select   id,  class,  scorefrom (  select        id,         class ,        score,        if(@cur_class <> class, @seq:=0, null) temp_seq,        if(@cur_class=class and @last_score <> score,@seq:=@seq+1,@seq) as seq ,        @cur_class:=class,        @last_score:=score    from         student a ,(select @cur_class:=0,@seq:=0, @last_score:=0) t0    order by         class asc , score desc) t1 where t1.seq <= 5;

解答2：使用子查询

select   t1.id,  t1.class,  t1.scorefrom   student t1where   (select count(1) c from student t2 where t2.class = t1.class and t2.score >= t1.score ) <= 5

前后记录关联——求次日留存

题目：如下表，记录用户访问流水，求次日留存率。

create table visit_log(  userId int(20) comment "用户id",  visitDate datetime comment "访问时间");

思路：次日留存率 = 交集(当天访问用户 和 下一天访问用户)/ 当天访问用户数

解答：

select    date_format(visitDate, "%Y-%m-%d") as `日期`,    count(distinct if(date_format(user_next_date, "%Y-%m-%d")) = date_format(date_add(visitDate,interval 1 day), "%Y-%m-%d"),uid , null ))     / count(distinct uid ) as  `次日留存率`from (    select        visitDate,         uid ,        -- 查出当前记录用户，下一次访问日期        (select min(visitDate) from visit_log where date_add(date_format(a.visitDate , "%Y-%m-%d 00:00:00"),interval 1 day) and a.userId = userId )     from         visit_log a ) a group by     date_format(visitDate, "%Y-%m-%d");