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
没有收藏的题目。

评论笔记

评论日期 题目名称 评论内容 站长评论
2024-11-07 一线城市历年平均气温  
哥后台验证代码好像没加年份限制
 
加了的，你再看看哦！
2024-11-06 表连接（3）一直使用一张表，现在开始两张表  
select mch_typ,count(mch_nm) as total_mch,count(distinct mch_nm) as unique_mch_cnt
from cmb_mch_typ 
group by mch_typ 
order by 2 desc;
那这个呢，请问还有什么问题啊o(╥﹏╥)o
 
刚看了后台，后台正确代码没加order by，现在加上了，你再试试，应该对了
2024-11-06 表连接（3）一直使用一张表，现在开始两张表  
select mch_typ,count(a.mch_nm) as total_mch,count(distinct a.mch_nm) as unique_mch_cnt 
from cmb_mch_typ a left join cmb_usr_trx_rcd b
on a.mch_nm=b.mch_nm 
group by mch_typ order by total_mch desc;
请问还有哪里有问题啊
 
看题干，先研究这张表，不需要join，一张表就够啦。实际业务中，也是先探索表、再join的
2024-10-30 分组与聚合函数（5）五花八门的项目，其实都有固定套路（2）  
陷阱挖哪了，找不到😭
 
我就说and or是个大坑吧，AND执行顺序高于or，3+2*2不等于10，等于7啊

提交记录

提交日期 题目名称 提交代码
2025-04-22 名字中字母e左起小于等于3位的歌手  
select singer_name,if(locate('e',singer_name)<=3 and locate('e',singer_name)>0 ,1,0) as if_e_lessthan3 
from singer_info;
2025-04-22 按歌手名字字符长度统计歌手个数  
select length(singer_name),count(singer_id) from singer_info group by 1;
2025-04-22 统计字符长度  
select singer_name,char_length(singer_name) as len from singer_info ;
2025-04-22 歌手名字大写  
select upper(singer_name) as uppered_name from singer_info ;
2025-03-17 经过第四象限的所有函数  
select * from numbers_for_fun
where (a=0 and ((b<0) or (b>=0 and c>0))) or (a>0) or (a<0 and c>0);
2025-03-17 经过第四象限的所有函数  
SELECT *
FROM numbers_for_fun
WHERE
    (a = 0 AND (
        (b = 0 AND c < 0) 
        OR (b != 0 AND (
            (b > 0 AND c < 0) 
            OR (b < 0)        
        ))
    ))
    OR
    (a != 0 AND (
        (a > 0 AND (c < 0 OR (b < 0 AND 4*a*c < b*b)))
        OR
        (a < 0)
    ));
2025-03-17 经过第四象限的所有函数  
select * from numbers_for_fun
where (a=0 and ((b<0) or (b>=0 and c>0))) or (a>0) or (a<0 and (b*b-4*a*c)<0);
2025-03-17 经过第四象限的所有函数  
select * from numbers_for_fun
where (a=0 and (b<0)or (b>=0 and c>0)) or a>0 or (a<0 and (b*b-4*a*c)<0);
2025-03-17 经过第四象限的所有函数  
select * from numbers_for_fun
where (a=0 and b>=0) or a>0 or (a<0 and (b*b-4*a*c)<0);
2025-03-17 经过第四象限的所有函数  
select * from numbers_for_fun
where (a=0 and b>=0) or (a!=0 and (b*b-4*a*c)>0);
2025-03-17 经过第四象限的所有函数  
select * from numbers_for_fun
where (a=0 and b>=0) or (b*b-4*a*c)>0;
2025-02-27 不经过第二象限的所有函数  
select * from numbers_for_fun 
where (a=0 and b>=0 and c<=0) or (a<0 and 
       (
           (b>=0 and c<=0)
       or (b<0 and c<=(b*b/4/a))
       )
       );
2025-02-27 不经过第二象限的所有函数  
select * from numbers_for_fun 
where (a=0 and b>0 and c>=-b) or (a<0 and 
       (
           (b>=0 and c<=0)
       or (b<0 and c<=(b*b/4/a))
       )
       );
2025-02-25 文科潜力股  
select * from scores 
where exam_date='2024-06-30' and subject in ('历史','政治','地理') and score>=90
order by score desc,student_id,subject;
2025-02-25 给英语成绩中上水平的学生拔尖  
select * from scores 
where exam_date='2024-06-30' and subject='英语' and score between 100 and 110 
order by score desc;
2025-02-25 找出三个班级的女生  
select * from students 
where class_code in('C219','C220','C221') and gender='f';
2025-02-25 大于J小于K的手牌  
select * from hand_permutations 
where card1 >'J' and card1<'K'
and card2>'J' and card2<'K';
2025-02-25 2000年以前出生的男歌手  
select * from singer_info 
where gender='m' and left(birth_date,4)<2000;
2025-02-25 国庆假期后第一天涨幅高于1%的股票  
select ts_code,open_price,close_price from daily_stock_prices 
where trade_date ='2023-10-09' and pct_change>1;
2025-02-25 总分超过300分的学生  
select student_id from subject_score 
where chinese+math+english>=300;