题目16： 同时在线问题
如下为某直播平台主播开播及关播时间，根据该数据计算出平台最高峰同时在线的主播人数。
id stt edt
1001,2021-06-14 12:12:12,2021-06-14 18:12:12
1003,2021-06-14 13:12:12,2021-06-14 16:12:12
1004,2021-06-14 13:15:12,2021-06-14 20:12:12
1002,2021-06-14 15:12:12,2021-06-14 16:12:12
1005,2021-06-14 15:18:12,2021-06-14 20:12:12
1001,2021-06-14 20:12:12,2021-06-14 23:12:12
1006,2021-06-14 21:12:12,2021-06-14 23:15:12
1007,2021-06-14 22:12:12,2021-06-14 23:10:12-- DDL
create table t16 (`id` string comment '主播id',`stt` string comment '主播登入时间',`edt` string comment '主播登出时间')comment '主播登入记录表'
row format delimited fields terminated by '\t'
lines terminated by '\n' stored as orc;insert overwrite table t16
select '1001', '2021-06-14 12:12:12', '2021-06-14 18:12:12' union all
select '1003', '2021-06-14 13:12:12', '2021-06-14 16:12:12' union all
select '1004', '2021-06-14 13:15:12', '2021-06-14 20:12:12' union all
select '1002', '2021-06-14 15:12:12', '2021-06-14 16:12:12' union all
select '1005', '2021-06-14 15:18:12', '2021-06-14 20:12:12' union all
select '1001', '2021-06-14 20:12:12', '2021-06-14 23:12:12' union all
select '1006', '2021-06-14 21:12:12', '2021-06-14 23:15:12' union all
select '1007', '2021-06-14 22:12:12', '2021-06-14 23:10:12' union all
select '1008', '2021-06-14 11:12:12', '2021-06-14 19:10:12' ;1. 思路分析1. 每条数据与本表自关联(笛卡尔积)2. 筛选出 同时在线的数据同时在线的条件 :1001 1----------52-----------6t1.stt <= t2.edtand t1.edt >= t2.stt-- 这个题目的思路是：每一个主播同时都跟谁同时在线。
-- 查询sql
with t as (select t1.id,t1.stt,t1.edt,count(distinct t2.id) onlineNum from t16 t1 left join t16 t2on t1.stt <= t2.edtand t1.edt >= t2.stt group by t1.id,t1.stt,t1.edt
)
select max(onlineNum) from t;-- 查询结果
t1.id t1.stt t1.edt cnt
1001 2021-06-14 12:12:12 2021-06-14 18:12:12 6
1001 2021-06-14 20:12:12 2021-06-14 23:12:12 5
1002 2021-06-14 15:12:12 2021-06-14 16:12:12 6
1003 2021-06-14 13:12:12 2021-06-14 16:12:12 6
1004 2021-06-14 13:15:12 2021-06-14 20:12:12 6
1005 2021-06-14 15:18:12 2021-06-14 20:12:12 6
1006 2021-06-14 21:12:12 2021-06-14 23:15:12 3
1007 2021-06-14 22:12:12 2021-06-14 23:10:12 3
1008 2021-06-14 11:12:12 2021-06-14 19:10:12 6说明 : 如下为某直播平台主播开播及关播时间，根据该数据计算出平台最高峰同时在线的主播人数-- 思路
-- 1. 将一条数据拆分成两条(id,dt,flag)，并且对数据进行打标：开播为1，关播为-1
-- 2. 然后按照dt排序，累计到当前时刻的flag之和，这个即为当前时刻主播在线的人数
-- 3. 获取上一步得到的主播在线人数的最大值，即为最高峰同时在线的主播人数-- 答案
with t as (selectid,stt dt , 1 flagfrom t16union allselectid,edt dt , -1 flagfrom t16
),t2 as (select id,dt,flag,sum(flag) over(order by dt) onlineNum from t
)
select max(onlineNum) from t2 ;题目17: 车辆每天上传N条数据，要求获得每日最后一条数据，请用至少三种方式写出来，并说明三种方式的区别
vin pkgts value
1111111111111111,2022-06-01 16:54:57,60.0
1111111111111111,2022-06-01 07:40:27,6.0
2222222222222222,2022-06-01 19:40:27,7.0
1111111111111111,2022-06-26 12:16:45,64.0
2222222222222222,2022-06-05 23:00:42,18.0
2222222222222222,2022-06-05 09:10:17,44.0
2222222222222222,2022-06-05 13:00:25,19.0
2222222222222222,2022-06-13 16:14:04,43.0
2222222222222222,2022-06-13 07:21:08,40.0-- 17三种方法第一种
row_number() 编号取最后
with s1 as
(select vin,split(pkgts,' ')[0] day,split(pkgts,' ')[1] times,value from t17),s2 as
(select *,row_number() over (partition by day order by times desc) key from s1)
select vin,concat(day,' ',times),value from s2 where key = 1;-- 17三种方法第二种
排序 用first_value 取分组内排序后，截止到当前行，第一个值 或 last_value 分组内排序后，截止到当前行，最后一个值
with s1 as
(select vin,split(pkgts,' ')[0] day,split(pkgts,' ')[1] times,value from t17),s2 as
(select last_value(day) over(partition by day order by times) k1,first_value(times) over(partition by day order by times desc) k2,first_value(vin) over(partition by day order by times desc ) vin,first_value(value) over(partition by day order by times desc ) valuefrom s1)
select distinct vin,concat(k1,' ',k2),value from s2 ;show create table t17-- 17三种方法第三种
用lead取下一天，要是最后一天就取不到为null把这列为null的全拿出来就是要的结果
with s1 as
(select *,split(pkgts,' ')[0] days from t17),s2 as
(select *,lead(pkgts) over (partition by days order by pkgts) key from s1)
select vin,pkgts,value from s2 where key is null;select version()
题目18：打折日期交叉问题
如下为平台商品促销数据：字段为品牌，打折开始日期，打折结束日期
brand stt edt
oppo,2021-06-05,2021-06-09
oppo,2021-06-11,2021-06-21
vivo,2021-06-05,2021-06-15
vivo,2021-06-09,2021-06-21
redmi,2021-06-05,2021-06-21
redmi,2021-06-09,2021-06-15
redmi,2021-06-17,2021-06-26
huawei,2021-06-05,2021-06-26
huawei,2021-06-09,2021-06-15
huawei,2021-06-17,2021-06-21计算每个品牌总的打折销售天数，注意其中的交叉日期，比如 vivo 品牌，
第一次活动时间为 2021-06-05 到 2021-06-15，第二次活动时间为 2021-06-09 到 2021-06-21 其中 9 号到 15号为重复天数，
只统计一次，即 vivo 总打折天数为 2021-06-05 到 2021-06-21 共计 17 天。with s1 as
(select *,lead(stt) over (partition by brand order by stt) nextstt,`if`(datediff(edt,lead(stt) over (partition by brand order by stt))>0,1,0) key from t18),s2 as
(select *,`if`(lag(key) over (partition by brand order by stt) = 1,'yes','no') finkey,row_number() over (partition by brand order by stt) numfrom s1),s3 as
(select *,lead(finkey) over (partition by brand order by stt) nextkey
from s2),s4 as
(select brand,`if`(num=1,`if`(nextkey='yes',datediff(lead(edt) over(partition by brand order by stt) ,stt)+1,datediff(edt,stt)+1),`if`(finkey='yes',0,datediff(edt,stt)+1)) n1from s3)
select brand,sum(n1) `打折时间` from s4 group by brand;
题目19: 【*】间隔连续问题
某游戏公司记录的用户每日登录数据
id dt
1001 2021-12-12
1002 2021-12-12
1001 2021-12-13
1001 2021-12-14
1001 2021-12-16
1002 2021-12-16
1001 2021-12-19
1002 2021-12-17
1001 2021-12-20计算每个用户最大的连续登录天数，可以间隔一天。解释：如果一个用户在 1,3,5,6 登录游戏，则视为连续 6 天登录。create table login_in(id int,dt string
)
row format delimited
fields terminated by ' ';load data local inpath '/home/hivedata/login.txt' into table login_in;select * from login_in;with t as (select *,lead(dt,1,dt) over (partition by id order by dt ) next_dt,if(datediff(lead(dt, 1, dt) over (partition by id order by dt ), dt) > 2, null,datediff(lead(dt, 1, dt) over (partition by id order by dt ), dt))daysfrom login_in
),t2 as (select *,sum(if(days <=2 ,0,1)) over (partition by id order by dt) groupId from t
),t3 as (select id,sum(days)+1 activeDays from t2 group by id,groupId
)
select id,max(activeDays) from t3 group by id;题目20：找出所有科目成绩都大于某一学科平均成绩的学生
表结构：uid,subject_id,score
数据集如下
1001 01 90
1001 02 90
1001 03 90
1002 01 85
1002 02 85
1002 03 70
1003 01 70
1003 02 70
1003 03 851）建表语句
create table score(
uid string,
subject_id string,
score int)
row format delimited fields terminated by '\t'; 2）求出每个学科平均成绩
select
uid,
score,
avg(score) over(partition by subject_id) avg_score
from
score;t13）根据是否大于平均成绩记录flag，大于则记为0否则记为1
select
uid,
if(score>avg_score,0,1) flag
from
t1;t24）根据学生id进行分组统计flag的和，和为0则是所有学科都大于平均成绩
select
uid
from
t2
group by
uid
having
sum(flag)=0;5）最终SQL
select
uid
from
(select
uid,
if(score>avg_score,0,1) flag
from
(select
uid,
score,
avg(score) over(partition by subject_id) avg_score
from
score)t1)t2
group by
uid
having
sum(flag)=0;
