Hive实际工作场景Sql题(业务自想)
HiveSql练习题
工作之余，结合业务所需构思的工作时常遇sql效果场景(实际业务场景可结合sql题自我构思)
有更好的sql解题思路欢迎大家到评论区交流
第一题
题目
数据原型:
time,t1,t2,t3
2021-07-01 00:01:01,1,4,1
2021-07-01 00:01:03,1,5,1
2021-07-01 00:01:11,1,6,1
2021-07-01 00:01:13,1,7,0
2021-07-01 00:01:23,1,8,0
2021-07-01 00:01:24,1,9,1
2021-07-01 00:02:24,1,10,1
所需效果:
2021-07-01 00:01:01,3,4,1,2021-07-01 00:01:01,2021-07-01 00:01:13
2021-07-01 00:01:13,2,7,0,2021-07-01 00:01:13,2021-07-01 00:01:24
2021-07-01 00:01:24,2,9,1,2021-07-01 00:01:24,2021-07-01 00:02:24
答案
答案方式1:
with tmp1 as(
select
time,
t1,t2,t3,
row_number() over(partition by t3 order by time) as sumOver
from demo10
),tmp2 as(
select
time,
t1,t2,t3,(t2-sumOver) as groupId
from tmp1
),tmp3 as (
select
min(time) as time,
min(t1) as tmpScheme,
sum(t1) as t1,
min(t2) as t2,
min(t3) as t3
from tmp2 group by groupId
),tmp4 as(
select
time,t1,t2,t3,
time,
lead(`time`,1,`time`) over(partition by tmpScheme order by `time` ) as last_time
from tmp3
)
select * from tmp4 order by `time`
答案方式2:
with tmp1 as(
select
time,
t1,t2,t3,(t2-t3) as diffNum,
sum(`t3`) over(partition by t1 order by time) as sumOver
from demo10
),tmp2 as(
select
time,t1,t2,t3,diffNum,sumOver,if(t3!=0,(diffNum-sumOver),(diffNum-sumOver)-diffNum) as groupId
from tmp1
),tmp3 as (
select
min(time) as time,
min(t1) as stationid,
sum(t1) as t1,
min(t2) as t2,
min(t3) as t3
from tmp2 group by groupId order by time
),tmp4 as (
select
time,t1,t2,t3,time as startTime,lead(time,1,time) over(partition by stationid order by time) as endTime
from tmp3
)
select * from tmp4
第二题
题目
数据原型
姓名,课程,分数
张三,语文,74
张三,数学,83
张三,物理,93
李四,语文,74
李四,数学,84
李四,物理,94
所需效果
姓名,语文,数学,物理,总分,平均分
李四,74,84,94,252,84.00
张三,74,83,93,250,83.33
答案
答案方式1
select
`name`,
sum(if(course='语文',score,0)) as chinese,
sum(if(course='数学',score,0)) as math,
sum(if(course='物理',score,0)) as physics,
sum(score) as sum,
round(avg(score),1) as avg
from questionV2 group by `name`
答案方式2
with tmp1 as (
select
`name`,
collect_list(cast(`score` as string))[0] as `chinese`,
collect_list(cast(`score` as string))[1] as `math`,
collect_list(cast(`score` as string))[2] as `physics`,
sum(`score`) as sum,
avg(`score`) as score
from questionV2 group by `name`
)
select * from tmp1
答案方式3
with tmp1 as (
select
`name`,
str_to_map(concat_ws(',',collect_set(concat_ws(':',course,score)))) as `courseList`,
sum(score) as sum,
avg(score) as avg
from
questionV2
group by `name`
),tmp2 as (
select
`name`,
courseList['语文'] as `chinese`,
courseList['数学'] as `math`,
courseList['物理'] as `physics`,
sum,avg
from tmp1
)
select * from tmp2
第三题
题目
数据原型
| stationid | hour1 | hour2 | hour3 | hour4 | hour5 | hour6 | hour7 | hour8 | hour9 | hour10 | hour11 | hour12 | hour13 | hour14 | hour15 | hour16 | hour17 | hour18 | hour19 | hour20 | hour21 | hour22 | hour23 | hour24 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| G01X01 | 11.0 | 12.0 | 13.0 | 14.0 | 15.0 | 16.0 | 17.0 | 18.0 | 19.0 | 20.0 | 21.0 | 22.0 | 23.0 | 24.0 | 25.0 | 26.0 | 27.0 | 28.0 | 29.0 | 30.0 | 31.0 | 32.0 | 33.0 | 34.0 |
展现效果
| stationid | col0 |
|---|---|
| G01X01 | 11.0 |
| G01X01 | 12.0 |
| G01X01 | 13.0 |
| G01X01 | 14.0 |
| G01X01 | 15.0 |
| G01X01 | 16.0 |
| G01X01 | 17.0 |
| G01X01 | 18.0 |
| G01X01 | 19.0 |
| G01X01 | 20.0 |
| G01X01 | 21.0 |
| G01X01 | 22.0 |
| G01X01 | 23.0 |
| G01X01 | 24.0 |
| G01X01 | 25.0 |
| G01X01 | 26.0 |
| G01X01 | 27.0 |
| G01X01 | 28.0 |
| G01X01 | 29.0 |
| G01X01 | 30.0 |
| G01X01 | 31.0 |
| G01X01 | 32.0 |
| G01X01 | 33.0 |
| G01X01 | 34.0 |
答案
实现方式1：
select
`stationid`,
posexplode(split(concat_ws(',',hour1,hour2,hour3,hour4,hour5,hour6,hour7,hour8,hour9,hour10,hour11,
hour12,hour13,hour14,hour15,hour16,hour17,hour18,hour19,hour20,hour21,hour22,hour23,hour24),","))
from questionV3
实现方式2：
select
`stationid`,
stack(24,hour1,hour2,hour3,hour4,hour5,hour6,hour7,hour8,hour9,hour10,hour11,hour12,hour13,hour14,
hour15,hour16,hour17,hour18,hour19,hour20,hour21,hour22,hour23,hour24)
from questionV3
题目四
题目
题目原型(前后数值超过5的 过滤出来)
| stationid | num | ct |
|---|---|---|
| A001 | 5 | 1623733736 |
| A001 | 10 | 1623733737 |
| A001 | 15 | 1623733738 |
| A001 | 20 | 1623733739 |
| A001 | 30 | 1623733740 |
| A001 | 35 | 1623733741 |
| A001 | 40 | 1623733742 |
| A001 | 50 | 1623733743 |
| A001 | 55 | 1623733744 |
展现效果
| stationid | num | dt | session_id |
|---|---|---|---|
| A002 | 5 | 1623733736 | 0 |
| A002 | 10 | 1623733737 | 0 |
| A002 | 15 | 1623733738 | 0 |
| A002 | 20 | 1623733739 | 0 |
| A002 | 30 | 1623733740 | 30 |
| A002 | 35 | 1623733741 | 30 |
| A002 | 40 | 1623733742 | 30 |
| A002 | 50 | 1623733743 | 50 |
| A002 | 55 | 1623733744 | 50 |
答案
实现方式1:
with tmp1 as(
select
uid,num,dt,
lag(`num`,1,0) over(partition by `uid` order by dt) as lag_num
from questionV4
),tmp2 as(
select uid,num,dt,lag_num,(num-lag_num) as session_id from tmp1
),tmp3 as(
select
uid,num,dt,
if(session_id>5,num,0) as max_session_id
from tmp2
),tmp4 as(
select
uid,num,dt,
max(`max_session_id`) over(partition by `uid` order by dt) as new_session_id
from tmp3
)
select * from tmp4