-- 數(shù)據(jù)準(zhǔn)備
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) 
)

-- 1. 對(duì)用戶(hù)數(shù)據(jù)進(jìn)行分組,按照活躍日期進(jìn)行排序(去重:防止有一天有多次活躍記錄)
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. 使用活躍日期和排序rn進(jìn)行差值計(jì)算,得到的日期如果是相等的,就說(shuō)明活躍日期是連續(xù)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 進(jìn)行分組求和,篩選出連續(xù)登陸天數(shù)大于3天的用戶(hù)
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
; 

-- 1. 將active_date 上抬2行,不存在默認(rèn)為'0'(計(jì)算連續(xù)活躍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躍記錄)
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. 過(guò)濾篩選出, lead_active_date 與 active_date 差值為2的, 差值2 -> 連續(xù)活躍了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, 得到連續(xù)活躍天數(shù)>=3天的用戶(hù)
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