180.1. Log in continuously for n days (database)

The original table

Use lag&lead+datediff Window function

An example of a miscalculation

SELECT a.user_id, a.login_date AS day1, LEAD(a.login_date,1)OVER(ORDER BY a.user_id) AS day2, LEAD(a.login_date,2)OVER(ORDER BY a.user_id) AS day3 FROM login_log a	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

Last time I was looking for consecutive numbers , You can find consecutive dates by changing them , Just use one more datediff Function is OK .
But what's different this time is , Each date has a corresponding user . Although the dates are continuous , If the date does not belong to the same user , Then it can't be counted .
chart 1 Because there is no grouping , direct order by, The result will be id by 2 Of users are also counted

SQL Code

Log in for three days , The output is 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 

--  Have a hybrid 
 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 

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

Log in for two consecutive days , The output is 1 and 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 

Use date_sub function

I use it msql, Different databases have different syntax

SELECT DATE_ADD('2022-02-21', INTERVAL 12 DAY),   DATE_SUB('2022-02-23', INTERVAL 13 DAY)
>> Output results 
2022-03-05			2022-02-10

SELECT user_id, login_date, RANK()OVER(PARTITION BY user_id ORDER BY login_date) rk FROM login_log	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

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;