以下是解决方案：

SELECT user_id, 
   DATEDIFF(MAX(order_time), MIN(order_time)) as order_diff
FROM orders
GROUP BY user_id
   HAVING order_diff > 0;

下面是一个测试它的链接。

以下方法可行：
架构（mysql v5.7）

CREATE TABLE orders
    (`user_id` int, `order_id` int, `order_time` datetime)
;

INSERT INTO orders
    (`user_id`, `order_id`, `order_time`)
VALUES
(1,208965785,'2016-12-15 17:14:13'),
(1,201765785,'2016-12-14 17:19:05'),
(1,203932785,'2016-12-13 20:41:30'),
(1,209612785,'2016-12-14 20:14:32'),
(1,208112785,'2016-12-14 20:27:08'),
(1,205525785,'2016-12-14 17:01:26'),
(1,208812785,'2016-12-14 20:18:23'),
(1,206432785,'2016-12-11 20:32:20'),
(1,206698785,'2016-12-14 10:50:15'),
(2,209524795,'2016-11-26 18:06:21'),
(3,206529925,'2016-10-01 10:43:57'),
(3,203729925,'2016-10-08 10:43:11'),
(4,204876145,'2016-09-24 10:23:49'),
(5,203363157,'2016-07-13 23:56:43'),
(6,207784875,'2017-01-04 12:21:21'),
(7,206437177,'2016-06-25 02:40:33'),
(8,202819645,'2016-09-09 11:47:27'),
(8,202819645,'2016-09-09 11:47:27'),
(8,202819646,'2016-09-08 11:47:27'),
(9,205127187,'2016-06-05 22:21:18'),
(9,205127187,'2016-06-05 22:21:18'),
(11,207874877,'2016-06-17 16:49:44'),
(12,204927595,'2016-11-28 23:05:40');

查询#1

SELECT dt2.user_id, 
       MIN(datediff(dt2.latest_order_time, 
                dt2.second_latest_order_time)) AS order_diff 
FROM (
 SELECT o.user_id, 
        o.order_time AS latest_order_time,  
        (SELECT o2.order_time 
         FROM orders AS o2 
         WHERE o2.user_id = o.user_id AND 
               o2.order_id <> o.order_id 
         ORDER BY o2.order_time DESC LIMIT 1) AS  second_latest_order_time 
 FROM orders AS o 
 JOIN (SELECT user_id, MAX(order_time) AS latest_order_time 
       FROM orders 
       GROUP BY user_id) AS dt 
   ON dt.user_id = o.user_id AND 
      dt.latest_order_time = o.order_time 
) AS dt2 
WHERE dt2.second_latest_order_time IS NOT NULL 
GROUP BY dt2.user_id;

| user_id | order_diff |
| ------- | ---------- |
| 1       | 1          |
| 3       | 7          |
| 8       | 1          |

db fiddle视图
细节：
我们确定最大值 order_time 为了一个 user_id 在子选择查询（派生表）中。我们可以把它化名为 latest_order_time .
我们 Join 此结果设置为 orders table。这将帮助我们只考虑最大值为的行 order_time 为了一个 user_id .
现在，我们使用相关子查询来确定最大 order_time 相同用户的值，超出 order_id 值。我们可以把它化名为 second_latest_order_time .
最后，再次将其用作派生表，并删除 second_latest_order_time 是 null ，并计算 datediff() 剩下的。
决赛 Group By 是必需的，因为您的数据有多个条目

根据你的小提琴：

select user_id, 
   datediff(max(order_time), 
            ( -- Scalar Subquery to get the 2nd largest order_time
              select max(order_time)
              from orders as o2
              where o2.user_id = o.user_id              -- same user
                and o2.order_time < max(o.order_time)   -- but not the max time
            )
           ) as diff
from orders as o
group by user_id
having diff is not null -- if there's no 2nd largest time diff will be NULL