Hive-SQL查詢連續活躍登錄用戶思路詳解

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連續活躍登陸的用戶指至少連續2天都活躍登錄的用戶

解決類似場景的問題

創建數據

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、因為每天用戶登錄次數可能不止一次，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 _)函數將用戶id分組，按照登陸時間進行排序。

3、計算登錄日期減去第二步驟得到的結果值，用戶連續登陸情況下，每次相減的結果都相同。

4、按照id和日期分組並求和，篩選大於等於2的即為連續活躍登陸的用戶。

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()函數計數

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：日期減去計數值得到結果

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id和結果分組並計算總和，大於等於2的即為連續登陸的用戶，得到 用戶id，開始日期，結束日期，連續登錄天數

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 結束日期 連續登錄天數

最後：連續登陸的用戶

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(向後)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-數據倉庫篇

SQL 查詢連續登陸7天以上的用戶

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