SQL统计连续登陆3天的用户(连续活跃超3天用户)_sql连续登录3天用户数

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

-- 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) 
)
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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
; 
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-- 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
; 
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-- 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
; 
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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
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-- 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
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-- 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
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end