题解 | #今天的刷题量(一)#

该题真的是我写SQL遇到的一个需求，整理了一个小问题出给大家做一下，比较固定的办法就是，写今天的日期，代码如下：

select subject.name,T.cnt from
(
    select subject_id,count(id) as cnt from  submission where `create_time` BETWEEN  '2022-02-23' AND  '2022-02-24'  group by subject_id
) as T join subject
on T.subject_id=subject.id
order by T.cnt desc ,subject.id;

当然我们也可以学习"今天"这个东西的用法，如下：

mysql
sqlite

select subject.name,T.cnt from
(
    select subject_id,count(id) as cnt from  submission where `create_time` BETWEEN  curdate() AND  DATE_ADD(curdate(),INTERVAL 1 DAY)  group by subject_id
) as T join subject
on T.subject_id=subject.id
order by T.cnt desc ,subject.id;
}

select subject.name,T.cnt from
(
    select subject_id,count(id) as cnt from  submission where `create_time` BETWEEN  curdate() AND  DATE_ADD(curdate(),INTERVAL 1 DAY)  group by subject_id
) as T join subject
on T.subject_id=subject.id
order by T.cnt desc ,subject.id;