Hive-SQL查詢連續(xù)活躍登錄用戶思路詳解
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(wèn)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(kāi)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(kāi)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(cāng)庫(kù)篇
SQL 查詢連續(xù)登陸7天以上的用戶
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