题解 | #今天的刷题量(一)#
今天的刷题量(一)
https://www.nowcoder.com/practice/e18f56796ae94c3d885e61c8d57a950e
select name,cnt -- 不需要subject_id
from (
select subject_id,s1.name,count(subject_id) cnt
from subject s1
join submission s2 on s1.id=s2.subject_id
where create_time=current_date -- 获取当日记录
group by subject_id,s1.name
order by cnt desc,subject_id -- 先按cnt降序，再按subject_id升序
)t;
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