排名

用户解题统计

过去一年提交了

勋章 ①金银铜：在竞赛中获得第一二三名；②好习惯：自然月10天提交；③里程碑：解决1/2/5/10/20/50/100/200题；④每周打卡挑战：完成每周5题，每年1月1日清零。

收藏

收藏日期 题目名称 解决状态
2025-09-10 天王天后的发烧友  未解决
2025-04-30 字符串与通配符（2）好多关键词做规则，可以使用rlike  已解决
2025-04-27 一线城市历年平均气温  已解决
2025-04-26 HAVING-每次成绩都不低于80分的学生  已解决

评论笔记

评论日期 题目名称 评论内容 站长评论
没有评论过的题目。

提交记录

提交日期 题目名称 提交代码
2025-09-10 接收红包金额绿茶榜  
select
	rcv_usr_id, sum(pkt_amt) as sum_trx_amt
from tx_red_pkt_rcd
where rcv_datetime != '1900-01-01 00:00:00'
group by rcv_usr_id
order by sum_trx_amt desc
limit 10;
2025-09-10 红包金额土豪榜  
select 
	snd_usr_id, sum(pkt_amt) as sum_trx_amt
from tx_red_pkt_rcd
group by snd_usr_id
order by sum_trx_amt desc
limit 10
;
2025-09-10 总分超过300分的学生  
select student_id
from subject_score
where math+english+chinese >= 300
order by student_id asc
2025-09-10 至少两门科目大于等于110分的学生  
with temp as 
(select
	student_id,
case when chinese >= 110 Then 1 Else 0 END as c,
case when math >= 110 Then 1 Else 0 END as m,
case when english >= 110 Then 1 Else 0 END as e
from subject_score s),
temp1 as
( select
 	student_id
 from temp 
 where c+m+e >=2
)
select *
from subject_score
where student_id in (select student_id from temp1)
order by student_id asc
2025-05-02 分组与聚合函数（2）擦边营收怎么样，聚合函数可看出  
select date(trx_time) as trx_time,
		max(trx_amt)as max_trx_amt,
min(trx_amt) as min_trx_amt,
avg(trx_amt) as avg_trx_amt,
sum(trx_amt) as total_trx_amt
from cmb_usr_trx_rcd
where year(trx_time) = 2024 and month(trx_time) = 9
	andmch_nm = "红玫瑰按摩保健休闲"
group by mch_nm,date(trx_time)
order by date(trx_time)
2025-04-30 开口向上且经过原点的一元二次函数  
select * from numbers_for_fun
where a>0 and c=0
2025-04-30 开口向上的一元二次函数  
select * from numbers_for_fun
where a>0
2025-04-30 必过(-1, -1)的一元一次函数  
select * from numbers_for_fun
where a=0 and b!=0 and (1-b+c=0)
2025-04-30 必过(0, 1)的一元一次函数  
select * from numbers_for_fun
where a=0 and b!=0 and c=1
2025-04-30 不经过第二象限的一元一次函数  
select * from numbers_for_fun
where a=0 and b>0 and c<=0
2025-04-30 不经过第三象限的一元一次函数  
select * from numbers_for_fun
where a=0 and b<0 and c>=0
2025-04-30 找出与y=x有交点的所有一元一次函数  
select * from numbers_for_fun
where a=0 and b!=0 and b!=1
2025-04-30 找出与y=x有交点的所有一元一次函数  
select * from numbers_for_fun
where a=0 and b<0
2025-04-30 找出与X轴交点小于等于0的一元一次函数  
select * from numbers_for_fun 
where a=0 and b!=0 and -c/b <=0
2025-04-30 找出与X轴交点大于0的一元一次函数  
select * from numbers_for_fun
where a=0 and b!=0 and -c/b >0
2025-04-30 找出所有经过原点的一元一次函数  
select * from numbers_for_fun
where a=0 and b!=0 and c=0
2025-04-30 找出所有一元一次函数  
select * from numbers_for_fun
where a=0 and b!=0
order by id asc;
2025-04-30 找出所有一元一次函数  
select * from numbers_for_fun
where a=1 and b!=0
order by id asc;
2025-04-30 性别已知的听歌用户  
select * from qqmusic_user_info
where gender in ('f','m') and year(birth_date) = 1980
order by birth_date asc
2025-04-30 性别已知的听歌用户  
select * from qqmusic_user_info
where gender != 'u' and year(birth_date) = 1980
order by birth_date asc