此查询对于计算每个用户的活动日期之间的最大间隔是否有效？

Question

I have the following task: Write a query to get the max gap between active dates for each user in the following table: (eg, user 1 will have max. gap of 3 days, user 2 will have 30 days, and user 3 will have null.)

CREATE TABLE IF NOT EXISTS `gamers` (
`id` int(6) not null,
`user_id` int(3) not null,
`active_date` DATE Not null,
  PRIMARY KEY (`id`)
  );
 INSERT INTO `gamers` (`id`,`user_id`,`active_date`) VALUES
  ('1','1','2019-01-01'),
  ('2','1','2019-01-02'),
  ('3','1','2019-01-05'),
  ('4','2','2019-03-01'),
  ('5','2','2019-03-31'),
  ('6','3','2019-04-01');

My solution would be the following:

SELECT g.id as id, g.user_id, MAX(total_amount_spent) OVER(PARTITION BY g.country), g.country
 FROM gamers2 as g
 INNER JOIN cte as c
 ON c.country = g.country
 WHERE install_source = 'ua' AND g.id NOT IN (SELECT id FROM cte)
 GROUP BY g.country
 ORDER BY g.user_id, total_amount_spent DESC
 ),
 cte3 AS(
 SELECT g.id, g.user_id, MAX(total_amount_spent) OVER(PARTITION BY g.country), g.country
 FROM gamers2 as g
 INNER JOIN cte2 as c
 ON c.country = g.country
 WHERE install_source = 'ua' AND g.id NOT IN (SELECT id FROM cte2)
 GROUP BY g.country
 ORDER BY g.user_id, total_amount_spent DESC
 )
 SELECT * FROM cte
 UNION
 SELECT * FROM cte2
 UNION
 SELECT * FROM cte3

Answer 1

We can use LAG() combined with ROW_NUMBER() as follows:

WITH cte AS (
    SELECT *, LAG(active_date, 1, active_date) OVER
                  (PARTITION BY user_id ORDER BY active_date) lag_active_date
    FROM gamers2
),
cte2 AS (
    SELECT *, DATEDIFF(active_date, lag_active_date) AS diff,
              ROW_NUMBER() OVER (PARTITION BY user_id
                                 ORDER BY DATEDIFF(active_date, lag_active_date) DESC) AS rn
    FROM cte
)

SELECT user_id, diff
FROM cte2
WHERE rn = 1;

Answer 2

A bit more readable, in my opinion:

WITH
w_lag_val AS (
  SELECT
    id
  , user_id
  , active_date
  , active_date 
    - LAG(active_date) OVER(
        PARTITION BY user_id ORDER BY active_date
      ) 
    AS lag_val
  FROM gamers
)
SELECT
  user_id
, MAX(lag_val) AS max_gap
FROM w_lag_val
GROUP BY user_id
ORDER BY user_id
;
-- out  user_id | max_gap 
-- out ---------+---------                                                                                                                                                                                 
-- out  1       |       3
-- out  2       |      30
-- out  3       |  (null)       

Answer 3

select *, max(diff) from (
select *, DATEDIFF(  f,active_date)as diff from (
SELECT gamers.id , gamers.user_id, gamers.active_date, 
(select active_date from gamers as g1 where g1.active_date > gamers.active_date and gamers.user_id = g1.user_id  order by g1.active_date limit 1) as f
from gamers
order by gamers.id) as vv) as gg
group by user_id

there is another solution