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SQL案例 - 统计用户获得积分

描述

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)

info_id user_id visit_time leave_time
911 10 2022-09-01 08:00:00 2022-09-01 09:02:00
912 11 2022-09-01 08:30:00 2022-09-01 09:10:00
913 12 2022-09-01 09:50:00 2022-09-01 10:12:00
914 13 2022-09-01 11:40:00 2022-09-01 12:22:00
921 11 2022-09-02 10:30:00 2022-09-02 11:05:00
922 11 2022-09-02 12:00:00 2022-09-02 12:02:00
923 12 2022-09-02 11:40:00 2022-09-02 13:15:00
924 13 2022-09-02 09:00:00 2022-09-02 09:02:00
925 14 2022-09-02 10:00:00 2022-09-02 10:40:00
931 10 2022-09-03 09:00:00 2022-09-03 09:22:00
932 11 2022-09-03 08:30:00 2022-09-03 09:10:00
933 13 2022-09-03 09:00:00 2022-09-03 09:32:00

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:

user_id point
11 11
12 11
10 8
13 7
14 4

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

答案

解法1:

select
    t.user_id,
    sum(
        floor(
            timestampdiff (minute, t.visit_time, t.leave_time) / 10
        )
    ) point
from
    visit_tb t
group by
    t.user_id
order by
    point desc;

*本案例来自牛客网,但答案为原创,如有雷同纯属巧合*

未经允许不得转载:夕枫 » SQL案例 - 统计用户获得积分
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