HiveSQL经典面试题（建议点赞收藏）_业界新闻

发布时间:2024-07-17 18:58

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经典面试系列

每科成绩都大于80分的学生信息

建表+初始化

--创建表scdn_student_score_test 并且初始化三个学生成绩 create table hdw_tmp_dev.scdn_student_score_test as  select '张三' as name,'数学' as subject ,'80' as score  union all  select '张三' as name,'语文' as subject ,'90' as score union all  select '张三' as name,'英语' as subject ,'90' as score  union all  select '李四' as name,'数学' as subject ,'90' as score union all  select '李四' as name,'语文' as subject ,'90' as score union all  select '李四' as name,'英语' as subject ,'70' as score  union all  select '王五' as name,'数学' as subject ,'90' as score union all  select '王五' as name,'语文' as subject ,'90' as score union all  select '王五' as name,'英语' as subject ,'50' as score  --查询结果显示 | scdn_student_score_test.name  | scdn_student_score_test.subject  | scdn_student_score_test.score  | +-------------------------------+----------------------------------+--------------------------------+ | 张三                            | 数学                               | 80                             | | 张三                            | 语文                               | 90                             | | 张三                            | 英语                               | 90                             | | 李四                            | 数学                               | 90                             | | 李四                            | 语文                               | 90                             | | 李四                            | 英语                               | 70                             | | 王五                            | 数学                               | 90                             | | 王五                            | 语文                               | 90                             | | 王五                            | 英语                               | 50                             | +-------------------------------+----------------------------------+---------------------  --求所有学科成绩都大于等于80分的学生姓名

思路一（思维转换）：所有问题找最小（最大）

select      t1.name     ,t1.min_score  from      (         select              name              ,min(score)  as min_score         from hdw_tmp_dev.scdn_student_score_test         group by name      ) as t1 --求出最小的成绩 where t1.min_score >=80 --最小成绩大于等于80，则这个学生的所有成绩都会大于等于80

执行结果

思路二（巧用左关联进行筛选）：

select      t1.name  from  (     select          name      from hdw_tmp_dev.scdn_student_score_test     group by name  ) as t1  left join  (     select          name      from hdw_tmp_dev.scdn_student_score_test     where score <80     group by name  ) as t2 on t1.name = t2.name where t2.name is null

连续登录问题

建表+初始化

create table hdw_tmp_dev.csdn_user_login_test as  select 'xiaoming' as user_name,'2024-01-01' as login_date  union all  select 'xiaoming' as user_name,'2024-01-02' as login_date  union all  select 'xiaoming' as user_name,'2024-01-03' as login_date union all  select 'xiaoming' as user_name,'2024-01-04' as login_date union all  select 'xiaoming' as user_name,'2024-01-05' as login_date union all  select 'dahuang' as user_name,'2024-01-02' as login_date  union all  select 'dahuang' as user_name,'2024-01-03' as login_date union all  select 'dahuang' as user_name,'2024-01-04' as login_date union all  select 'dahuang' as user_name,'2024-01-05' as login_date union all  select 'lucky_dog' as user_name,'2024-01-01' as login_date  union all  select 'lucky_dog' as user_name,'2024-01-03' as login_date union all  select 'lucky_dog' as user_name,'2024-01-04' as login_date union all  select 'lucky_dog' as user_name,'2024-01-05' as login_date

Sql参考

select      t2.user_name      ,t2.date_begin_flag     ,count(1) as max_login_days from  (     select          t1.user_name          ,t1.login_date         ,date_sub(date(login_date),t1.rn) as date_begin_flag     from      (         select              user_name              ,login_date              ,row_number()over(partition by user_name order by login_date) as rn          from hdw_tmp_dev.csdn_user_login_test     ) as t1 ) as t2   group by t2.user_name      ,t2.date_begin_flag

执行过程+结果

行列转换问题

建表+初始化：参考问题一的表

| t1.name              | t1.subject             | t1.score  | +-------------------------------+----------------------------------+-----------------+ | 张三                 | 数学                    | 80                             | | 张三                 | 语文                    | 90                             | | 张三                 | 英语                    | 90                             | | 李四                 | 数学                    | 90                             | | 李四                 | 语文                    | 90                             | | 李四                 | 英语                    | 70                             | | 王五                 | 数学                    | 90                             | | 王五                 | 语文                    | 90                             | | 王五                 | 英语                    | 50                             | +-------------------------------+----------------------------------+------------------

行专列Sql参考

select      name      ,max(case when subject = '数学' then score end) as math_score     ,max(case when subject = '语文' then score end) as china_score     ,max(case when subject = '英语' then score end) as english_score from hdw_tmp_dev.scdn_student_score_test group by name

行专列执行结果

列转行：可以参考建表初始化语句（union all）

select      name     ,collect_set(subject)  as subject_set from hdw_tmp_dev.scdn_student_score_test group by name

每个学生选课结果（多行变一行）：

将上面的结果展开（一行变多行）数据准备：

create table hdw_tmp_dev.scdn_student_score_test_collect as  select 'zhangsan' as name ,'"数学","语文","英语"' as subject_list union all  select 'lisi' as name ,'"美术","生物","物理"' as subject_list union all  select 'wangwu' as name ,'"计算机","日语","韩语"' as subject_list

炸开代码

select      name     ,subject_list     ,subject_name from hdw_tmp_dev.scdn_student_score_test_collect lateral view explode(split(subject_list,',')) extend_sub as subject_name

结果

留存问题：看当天登录后第N天是否登录

建表+初始化：参照连续登录表

Sql参考

--方案一：利用lead(日期，N)是否等于 当天登录实践+N天 select      t1.user_name     ,t1.logon_date     ,case when lead1_logon_date = date_add(logon_date,1) then '1天留存' end as 1day_remain     ,case when lead3_logon_date = date_add(logon_date,3) then '3天留存' end as 3day_remain from  (     select          user_name          ,logon_date          ,lead(user_name,1)over(partition by user_name order by logon_date) as lead1_user_name         ,lead(logon_date,1)over(partition by user_name order by logon_date) as lead1_logon_date         ,lead(user_name,3)over(partition by user_name order by logon_date) as lead3_user_name         ,lead(logon_date,3)over(partition by user_name order by logon_date) as lead3_logon_date     from hdw_tmp_dev.csdn_user_logon_test ) as t1  --方案二： select      t2.first_log_date       as first_log_date     ,count(t2.user_id)      as new_user_cnt --新用户数      ,count(t3.user_id)      as next_user_id --次日回访用户数     ,count(t4.user_id)      as 30_user_id   --30天回访用户数     ,count(t3.user_id)/count(t2.user_id)    as next_back_rate --次日回访率     ,count(t4.user_id)/count(t2.user_id)    as 30_back_rate   --30天回访率 from  (     select          first_log_date         ,user_id          ,date_add(first_log_date,1)  as next_log_date         ,date_add(first_log_date,29) as 30_log_date     from      (         select              user_id              ,log_time              ,first_value(date(log_time))over(partition by user_id) as first_log_date          from user_log      ) as t1      group by first_log_date         ,user_id  ) as t2  left join  (     select          user_id          ,date(log_date) as log_date       from user_log      group by user_id          ,date(log_date) as log_date  ) as t3 on t2.user_id = t2.user_id and t2.next_log_date = t3.log_date left join  (     select          user_id          ,date(log_date) as log_date       from user_log      group by user_id          ,date(log_date) as log_date  ) as t4 on t2.user_id = t4.user_id and t2.30_log_date = t4.log_date group by t2.first_log_date

方案一执行结果

TopN问题

建表+初始化

create table hdw_tmp_dev.scdn_student_score_test1 as  select '张三' as name,'数学' as subject ,'80' as score  union all  select '张三' as name,'语文' as subject ,'90' as score union all  select '张三' as name,'英语' as subject ,'90' as score  union all  select '李四' as name,'数学' as subject ,'90' as score union all  select '李四' as name,'语文' as subject ,'90' as score union all  select '李四' as name,'英语' as subject ,'70' as score  union all  select '王五' as name,'数学' as subject ,'90' as score union all  select '王五' as name,'语文' as subject ,'90' as score union all  select '王五' as name,'英语' as subject ,'50' as score  union all  select '小明' as name,'数学' as subject ,'88' as score union all  select '小明' as name,'语文' as subject ,'99' as score union all  select '小明' as name,'英语' as subject ,'77' as score union all  select '小文' as name,'数学' as subject ,'66' as score union all  select '小文' as name,'语文' as subject ,'89' as score union all  select '小文' as name,'英语' as subject ,'90' as score

Sql参考：求每科前三名对应的人员的成绩单

select      * from  (     select          name         ,subject         ,score         ,row_number()over(partition by subject order by score desc) as rn      from hdw_tmp_dev.scdn_student_score_test1 ) as t1  where rn<=3

执行结果

累计计算问题

建表+初始化

create table hdw_tmp_dev.user_sale_date as  select '001' as user_id,'2024-02-01' as sale_date, 100 as amount  union all  select '001' as user_id,'2024-02-02' as sale_date, 200 as amount  union all  select '001' as user_id,'2024-02-03' as sale_date, 300 as amount  union all  select '001' as user_id,'2024-02-04' as sale_date, 400 as amount  union all  select '001' as user_id,'2024-02-05' as sale_date, 500 as amount  union all  select '001' as user_id,'2024-02-06' as sale_date, 600 as amount

SQL逻辑

select      user_id     ,sale_date     ,amount     ,sum(amount)over(partition by user_id order by sale_date) as accuma_amount --按日期逐渐累加     ,sum(amount)over(partition by user_id) as total_amount --按人汇总     ,avg(amount)over(partition by user_id) as avg_amount --安人平均每天     ,max(amount)over(partition by user_id) as max_amount --单日最大销售     ,min(amount)over(partition by user_id) as min_amount --单日最小销售 from hdw_tmp_dev.user_sale_date

结果展示

HSQL进阶版

直播间在线最大人数

建表+初始化

create table hdw_tmp_dev.csdn_user_login_time_detail as  select '001' as user_id,'2024-02-01 10:00:00' as begin_time,'2024-02-01 12:00:00' as end_time  union all  select '002' as user_id,'2024-02-01 11:00:00' as begin_time,'2024-02-01 13:00:00' as end_time  union all  select '003' as user_id,'2024-02-01 11:00:00' as begin_time,'2024-02-01 14:00:00' as end_time  union all  select '004' as user_id,'2024-02-01 15:00:00' as begin_time,'2024-02-01 16:00:00' as end_time

SQL逻辑

select      t1.user_id      ,t1.time1     ,t1.flag     ,sum(flag)over(order by t1.time1) as user_cnt  from  (     select          user_id         ,begin_time as time1          ,1    as flag      from hdw_tmp_dev.csdn_user_login_time_detail     union all      select          user_id         ,end_time         ,-1    as flag      from hdw_tmp_dev.csdn_user_login_time_detail ) as t1

结果展示

SQL循环

建表+初始化

create table hdw_tmp_dev.cycle_1 as  select '1011' as a  union all  select '0101' as a

SQL逻辑

select           a,         concat_ws(",",collect_list(cast(index  as  string)))  as  res from  (         select                   a,                 index+1  as  index,                 chr         from  (                 select                           a,                         concat_ws(",",substr(a,1,1),substr(a,2,1),substr(a,3,1),substr(a,-1))  str                 from  hdw_tmp_dev.cycle_1 as t8         )  tmp1         lateral  view  posexplode(split(str,","))  t  as  index,chr         where  chr  =  "1" )  tmp2 group  by  a;

结果展示

计算中位数

建表+初始化:

create table hdw_tmp_dev.user_sale_date as  select '001' as user_id,'2024-02-01' as sale_date, 100 as amount  union all  select '001' as user_id,'2024-02-02' as sale_date, 200 as amount  union all  select '001' as user_id,'2024-02-03' as sale_date, 300 as amount  union all  select '001' as user_id,'2024-02-04' as sale_date, 400 as amount  union all  select '001' as user_id,'2024-02-05' as sale_date, 500 as amount  union all  select '001' as user_id,'2024-02-06' as sale_date, 600 as amount

SQL逻辑

select      t1.user_id     ,t1.sale_date     ,t1.amount     ,avg(t1.amount)over(partition by t1.user_id) as zhongwenshu from  (     select          user_id         ,sale_date         ,amount         ,row_number()over(partition by user_id order by amount) as rn      from hdw_tmp_dev.user_sale_date ) as t1  left join  (     select          user_id          ,count(1) as cnt      from hdw_tmp_dev.user_sale_date     group by user_id  ) as t2 on t1.user_id = t2.user_id  where t1.rn in (cnt/2,(cnt+1)/2,cnt/2+1) --总个数为奇数命中(cnt+1)/2；总个数为偶数命中：cnt/2,cnt/2+1，两数相加求平均值

结果展示

产生连续数值

SQL逻辑

--产生1到10的连续数据 select     start_id + pos  as  id     ,pos     ,val from(     select         1   as  start_id,         10  as  end_id )  m  lateral  view  posexplode(split(space(end_id - start_id),''))  t  as  pos,  val --方案二 select     row_number()  over()  as  id from         (select  split(space(99), '') as  x)  t lateral  view explode(x)  ex;