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统计用户获得积分

[编程题]统计用户获得积分
  • 热度指数:46 时间限制:C/C++ 1秒,其他语言2秒 空间限制:C/C++ 256M,其他语言512M
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现有某商家用户访问数据如下所示:
用户访问信息表:visit_tb(访问信息id-info_id,用户id-user_id,访问时间-visit_time,离开时间-leave_time)

该商城这几日推出新的推广活动,用户单次访问时长满10分钟则获得1积分,请查询这几日访问的用户可以获得多少积分?
要求输出:user_id,积分
注:输出结果按照积分降序排序
示例数据结果如下:
结果解释:
user_id为11的用户分别在0901 08:30、0902 10:30、0902 12:00、0903 08:30访问,
访问时长分别为40、35、2、40分钟,故获得积分为4+3+0+4=11积分;
其他结果同理
示例1

输入

drop table if exists  `visit_tb` ; 
CREATE TABLE `visit_tb` (
`info_id` int(11) NOT NULL,
`user_id` int(11) NOT NULL,
`visit_time` datetime NOT NULL,
`leave_time` datetime NOT NULL,
PRIMARY KEY (`info_id`));
INSERT INTO visit_tb VALUES(0911,10,'2022-09-01 08:00:00','2022-09-01 09:02:00'); 
INSERT INTO visit_tb VALUES(0912,11,'2022-09-01 08:30:00','2022-09-01 09:10:00'); 
INSERT INTO visit_tb VALUES(0913,12,'2022-09-01 09:50:00','2022-09-01 10:12:00'); 
INSERT INTO visit_tb VALUES(0914,13,'2022-09-01 11:40:00','2022-09-01 12:22:00'); 
INSERT INTO visit_tb VALUES(0921,11,'2022-09-02 10:30:00','2022-09-02 11:05:00'); 
INSERT INTO visit_tb VALUES(0922,11,'2022-09-02 12:00:00','2022-09-02 12:02:00'); 
INSERT INTO visit_tb VALUES(0923,12,'2022-09-02 11:40:00','2022-09-02 13:15:00'); 
INSERT INTO visit_tb VALUES(0924,13,'2022-09-02 09:00:00','2022-09-02 09:02:00'); 
INSERT INTO visit_tb VALUES(0925,14,'2022-09-02 10:00:00','2022-09-02 10:40:00'); 
INSERT INTO visit_tb VALUES(0931,10,'2022-09-03 09:00:00','2022-09-03 09:22:00'); 
INSERT INTO visit_tb VALUES(0932,11,'2022-09-03 08:30:00','2022-09-03 09:10:00'); 
INSERT INTO visit_tb VALUES(0933,13,'2022-09-03 09:00:00','2022-09-03 09:32:00'); 

输出

user_id|point
11|11
12|11
10|8
13|7
14|4
with t1 as (
    select *, timestampdiff(MINUTE, visit_time, leave_time) as duration
    from visit_tb
)

select user_id, sum(floor(duration/10)) as point
from t1
group by user_id
order by point desc

编辑于 2024-03-03 17:40:13 回复(0)
SELECT user_id,
       SUM(FLOOR(TIMESTAMPDIFF(MINUTE, visit_time, leave_time) / 10)) AS points
FROM visit_tb
GROUP BY user_id
ORDER BY points DESC;

编辑于 2024-04-07 10:10:22 回复(0)
select user_id, sum(point) point 
from (
    select user_id, FLOOR(timestampdiff(MINUTE, visit_time, leave_time )/ 10) point 
    from visit_tb
) t 
group by user_id 
order by sum(t.point) desc
发表于 2024-03-08 15:01:08 回复(0)