SQL统计连续登陆3天的用户(连续活跃超3天用户)

目录

• • SQL统计连续登陆3天的用户(连续活跃超3天用户)
  • • 1. 数据准备
    • 2. 方法一: 差值计算
    • 3. 方法二: lead或lag函数
    • • end

1. 数据准备

-- 数据准备 WITH user_active_info AS ( SELECT * FROM (     VALUES ('10001' , '2023-02-01'),('10001' , '2023-02-03')           ,('10001' , '2023-02-04'),('10001' , '2023-02-05')           ,('10002' , '2023-02-02'),('10002' , '2023-02-03')           ,('10002' , '2023-02-04'),('10002' , '2023-02-05')           ,('10002' , '2023-02-07'),('10003' , '2023-02-02')           ,('10003' , '2023-02-03'),('10003' , '2023-02-04')           ,('10003' , '2023-02-05'),('10003' , '2023-02-06')           ,('10003' , '2023-02-07'),('10003' , '2023-02-08')           ,('10004' , '2023-02-03'),('10004' , '2023-02-04')           ,('10004' , '2023-02-06'),('10004' , '2023-02-07')           ,('10004' , '2023-02-08'),('10004' , '2023-02-08')      	  ,('10005' , '2023-02-02'),('10005' , '2023-02-05')  ) AS user_active_info(user_id, active_date)  ) 

2. 方法一: 差值计算

-- 1. 对用户数据进行分组,按照活跃日期进行排序(去重:防止有一天有多次活跃记录) SELECT        user_id     , active_date     , ROW_NUMBER() OVER(PARTITION BY user_id ORDER BY active_date) AS rn  FROM user_active_info GROUP BY user_id , active_date ;  

-- 2. 使用活跃日期和排序rn进行差值计算,得到的日期如果是相等的,就说明活跃日期是连续的 SELECT        user_id     , active_date     , rn      , DATE_SUB(active_date,rn) AS sub_date FROM (     SELECT            user_id         , active_date         , ROW_NUMBER() OVER(PARTITION BY user_id ORDER BY active_date) AS rn      FROM user_active_info      GROUP BY user_id , active_date     ) a ;  

-- 3. 按照user_id和sub_date 进行分组求和,筛选出连续登陆天数大于3天的用户 SELECT        user_id     , MIN(active_date) AS begin_date     , MAX(active_date) AS end_date     , COUNT (1) AS login_duration FROM (     SELECT            user_id         , active_date         , rn          , DATE_SUB(active_date,rn) AS sub_date     FROM (         SELECT                user_id             , active_date             , ROW_NUMBER() OVER(PARTITION BY user_id ORDER BY active_date) AS rn          FROM user_active_info          GROUP BY user_id , active_date     ) a ) b GROUP BY user_id , sub_date HAVING login_duration >= 3 ;  

3. 方法二: lead或lag函数

-- 1. 将active_date 上抬2行,不存在默认为'0'(计算连续活跃3天以上的, 上抬2行,n天上抬n-1行)(去重:防止有一天有多次活跃记录) SELECT        user_id     , active_date     , lead(active_date , 2 , 0) OVER(PARTITION BY user_id ORDER BY active_date) AS lead_active_date  FROM user_active_info GROUP BY user_id , active_date 

-- 2. 过滤筛选出, lead_active_date 与 active_date 差值为2的, 差值2 -> 连续活跃了3天 SELECT        user_id , active_date , lead_active_date FROM (     SELECT            user_id         , active_date         , lead(active_date , 2 , 0) OVER(PARTITION BY user_id ORDER BY active_date) AS lead_active_date     FROM user_active_info     GROUP BY user_id , active_date ) a  WHERE  lead_active_date != '0'     AND DATEDIFF(lead_active_date , active_date) = 2 

-- 3. user_id 去重, 得到连续活跃天数>=3天的用户 SELECT        user_id FROM (     SELECT            user_id , active_date , lead_active_date     FROM (         SELECT                user_id             , active_date             , lead(active_date , 2 , 0) OVER(PARTITION BY user_id ORDER BY active_date) AS lead_active_date          FROM user_active_info         GROUP BY user_id , active_date     ) a      WHERE  lead_active_date != '0'         AND DATEDIFF(lead_active_date , active_date) = 2 ) b GROUP BY user_id 

end