Hive-SQL查詢連續(xù)活躍登錄用戶思路詳解

更新時(shí)間：2021年12月16日 14:08:57 作者：奇遇yms

這篇文章主要介紹了Hive-SQL查詢連續(xù)活躍登陸的用戶,活躍用戶這里是指連續(xù)2天都活躍登錄的用戶，本文給大家分享解決思路及sql語句，對(duì)大家的學(xué)習(xí)或工作具有一定的參考借鑒價(jià)值，需要的朋友可以參考下

連續(xù)活躍登陸的用戶指至少連續(xù)2天都活躍登錄的用戶

解決類似場(chǎng)景的問題

創(chuàng)建數(shù)據(jù)

CREATE TABLE test5active(
dt string,
user_id string,
age int)
ROW format delimited fields terminated BY ',';

INSERT INTO TABLE test5active VALUES 
('2019-02-11','user_1',23),('2019-02-11','user_2',19),
('2019-02-11','user_3',39),('2019-02-11','user_1',23),
('2019-02-11','user_3',39),('2019-02-11','user_1',23),
('2019-02-12','user_2',19),('2019-02-13','user_1',23),
('2019-02-15','user_2',19),('2019-02-16','user_2',19);

思路一:

1、因?yàn)槊刻煊脩舻卿洿螖?shù)可能不止一次，所以需要先將用戶每天的登錄日期去重。

2、再用row_number() over(partition by _ order by _)函數(shù)將用戶id分組，按照登陸時(shí)間進(jìn)行排序。

3、計(jì)算登錄日期減去第二步驟得到的結(jié)果值，用戶連續(xù)登陸情況下，每次相減的結(jié)果都相同。

4、按照id和日期分組并求和，篩選大于等于2的即為連續(xù)活躍登陸的用戶。

第一步：用戶登錄日期去重

select DISTINCT dt,user_id from test5active;

第二步：用row_number() over()函數(shù)計(jì)數(shù)

select 
t1.user_id,t1.dt,
row_number() over(partition by t1.user_id order by t1.dt) day_rank
from 
(
select DISTINCT dt,user_id from test5active
)t1;

第三步：日期減去計(jì)數(shù)值得到結(jié)果

select
t2.user_id,t2.dt,date_sub(t2.dt,t2.day_rank) as dis
from 
(
select 
t1.user_id,t1.dt,
row_number() over(partition by t1.user_id order by t1.dt) day_rank
from 
(
select DISTINCT dt,user_id from test5active
)t1)t2;

第四步：根據(jù)id和結(jié)果分組并計(jì)算總和，大于等于2的即為連續(xù)登陸的用戶，得到用戶id，開始日期，結(jié)束日期，連續(xù)登錄天數(shù)

select 
t3.user_id,min(t3.dt),max(t3.dt),count(1)
from
(
select
t2.user_id,t2.dt,date_sub(t2.dt,t2.day_rank) as dis
from 
(
select 
t1.user_id,t1.dt,
row_number() over(partition by t1.user_id order by t1.dt) day_rank
from 
(
select DISTINCT dt,user_id from test5active
)t1
)t2
)t3 group by t3.user_id,t3.dis having count(1)>1;

用戶id 開始日期結(jié)束日期連續(xù)登錄天數(shù)

最后：連續(xù)登陸的用戶

select distinct t4.user_id
from
(
select 
t3.user_id,min(t3.dt),max(t3.dt),count(1)
from
(
select
t2.user_id,t2.dt,date_sub(t2.dt,t2.day_rank) as dis
from 
(
select 
t1.user_id,t1.dt,
row_number() over(partition by t1.user_id order by t1.dt) day_rank
from 
(
select DISTINCT dt,user_id from test5active
)t1
)t2
)t3 group by t3.user_id,t3.dis having count(1)>1
)t4;

思路二：使用lag(向后)或者lead(向前)

select 
user_id,t1.dt,
lead(t1.dt) over(partition by user_id order by t1.dt) as last_date_id
from 
(
select DISTINCT dt,user_id from test5active
)t1;

select
distinct t2.user_id
from 
(
select 
user_id,t1.dt,
lead(t1.dt) over(partition by user_id order by t1.dt) as last_date_id
from 
(
select DISTINCT dt,user_id from test5active
)t1
)t2 where datediff(last_date_id,t2.dt)=1;

參考：

2020年大廠面試題-數(shù)據(jù)倉庫篇

SQL 查詢連續(xù)登陸7天以上的用戶

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