原表

使用lag&lead+datediff窗口函数

一个计算有误的例子

上次是求连续出现的数字，变换一下就能求连续的日期，只需要多用一个datediff函数就行。
但这次不一样的是，每一个日期都有与之对应的用户。虽然日期是连续的，倘若日期不属于同一个用户，那么就不能计算在内。
图1就是因为没分组，直接order by，导致输出结果将id为2的用户也计算在内

SQL代码

连续登录三天，输出结果是1

SELECT DISTINCT t.user_id FROM 
(
SELECT a.user_id, 
a.login_date AS day1,
LEAD(a.login_date,1)OVER(PARTITION BY a.user_id ORDER BY a.user_id) AS day2,
LEAD(a.login_date,2)OVER(PARTITION BY a.user_id ORDER BY a.user_id) AS day3 
FROM login_log a
) t 
WHERE DATEDIFF(t.day2,t.day1)=1 AND DATEDIFF(t.day3,t.day2)=1 

-- 来个混合型的
 SELECT DISTINCT user_id
  FROM
      (SELECT user_id,
            LAG(login_date,1) OVER(PARTITION BY user_id ORDER BY login_date) AS lag_login_date,
            login_date,
            LEAD(login_date,1) OVER(PARTITION BY user_id ORDER BY login_date) AS lead_login_date
      FROM dwd.login_log)t1
  WHERE DATEDIFF(login_date,lag_login_date)=1 AND DATEDIFF(lead_login_date,login_date)=1 
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SELECT user_id,
LAG(login_date,1) OVER(PARTITION BY user_id ORDER BY login_date) AS lag_login_date,
login_date,
LEAD(login_date,1) OVER(PARTITION BY user_id ORDER BY login_date) AS lead_login_date
FROM login_log

连续登录两天，输出结果是1和3

SELECT DISTINCT t.user_id FROM 
(
SELECT a.user_id, 
a.login_date AS day1,
LEAD(a.login_date,1)OVER(PARTITION BY a.user_id ORDER BY a.user_id) AS day2,
LEAD(a.login_date,2)OVER(PARTITION BY a.user_id ORDER BY a.user_id) AS day3 
FROM login_log a
) t 
WHERE DATEDIFF(t.day2,t.day1)=1 
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使用date_sub函数

我用的是msql，不同数据库之间语法会有差异

SELECT DATE_ADD('2022-02-21', INTERVAL 12 DAY),   DATE_SUB('2022-02-23', INTERVAL 13 DAY)
>>输出结果
2022-03-05			2022-02-10
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SELECT user_id,con_login_date,COUNT(*) nums
FROM
    (
    SELECT user_id,login_date,rk-1, DATE_SUB(login_date, INTERVAL t1.rk-1 DAY) AS con_login_date
    FROM 
        (SELECT user_id,login_date,RANK()OVER(PARTITION BY user_id ORDER BY login_date) rk
         FROM login_log) t1
    )t2
GROUP BY user_id,con_login_date
HAVING COUNT(*) >= 3;
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