SQL统计连续登陆3天的用户(连续活跃超3天用户)

SQL统计连续登陆3天的用户(连续活跃超3天用户)

目录

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(PARTITIONBY user_id ORDERBY active_date)AS rn 
FROM user_active_info
GROUPBY user_id , active_date
;

user_idactive_datern100012023-02-011100012023-02-032100012023-02-043100012023-02-054100022023-02-021100022023-02-032100022023-02-043100022023-02-054100022023-02-075………

-- 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(PARTITIONBY user_id ORDERBY active_date)AS rn 
    FROM user_active_info 
    GROUPBY user_id , active_date
    ) a
;

user_idactive_daternsub_date100012023-02-0112023-01-31100012023-02-0322023-02-01100012023-02-0432023-02-01100012023-02-0542023-02-01100022023-02-0212023-02-01100022023-02-0322023-02-01100022023-02-0432023-02-01100022023-02-0542023-02-01100022023-02-0752023-02-02…………

-- 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(PARTITIONBY user_id ORDERBY active_date)AS rn 
        FROM user_active_info 
        GROUPBY user_id , active_date
    ) a
) b
GROUPBY user_id , sub_date
HAVING login_duration >=3;

user_idbegin_dateend_datelogin_duration100012023-02-032023-02-053100022023-02-022023-02-054100032023-02-022023-02-087100042023-02-062023-02-083

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(PARTITIONBY user_id ORDERBY active_date)AS lead_active_date 
FROM user_active_info
GROUPBY user_id , active_date

user_idactive_datelead_active_date100012023-02-012023-02-04100012023-02-032023-02-05100012023-02-040100012023-02-050100022023-02-022023-02-04100022023-02-032023-02-05100022023-02-042023-02-07100022023-02-050100022023-02-070………

-- 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(PARTITIONBY user_id ORDERBY active_date)AS lead_active_date
    FROM user_active_info
    GROUPBY user_id , active_date
) a 
WHERE  lead_active_date !='0'AND DATEDIFF(lead_active_date , active_date)=2

user_idactive_datelead_active_date100012023-02-032023-02-05100022023-02-022023-02-04100022023-02-032023-02-05………

-- 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(PARTITIONBY user_id ORDERBY active_date)AS lead_active_date 
        FROM user_active_info
        GROUPBY user_id , active_date
    ) a 
    WHERE  lead_active_date !='0'AND DATEDIFF(lead_active_date , active_date)=2) b
GROUPBY user_id

user_id10001100021000310004

end