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
2025-05-21 多云天气天数  已解决
2025-05-21 字符串与通配符(2)好多关键词做规则,可以使用rlike  已解决
2025-05-10 分类(1)姿势太多很过分,分类要用CaseWhen  已解决

评论笔记

评论日期 题目名称 评论内容 站长评论
2025-05-21 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

group by 1 什么意思,怎么理解?
第一列,简写

提交记录

提交日期 题目名称 提交代码
2025-05-25 学生信息和班主任姓名 
select 
	s.name,
s.class_code,
s.grade_code,
t.name as head_teacher_name
from students s
join teachers t on s.class_code = t.head_teacher
order by 
	s.student_id
2025-05-21 按歌手名字字符长度统计歌手个数 
select length(singer_name),count(1)
from singer_info
group by
	length(singer_name)
2025-05-21 统计字符长度 
select singer_name ,char_length(singer_name)
from singer_info
2025-05-21 歌手名字大写 
select upper(singer_name)
from singer_info
2025-05-21 北京有雪的日子 
select dt,tmp_h,tmp_l,con as wnd from weather_rcd_china
where con like '%雪%' and lower(city) = 'beijing'
2025-05-21 北京有雪的日子 
select dt,tmp_h,tmp_l,con as wnd from weather_rcd_china
where con like '%雪%'
2025-05-21 北京有雪的日子 
select dt,tmp_h,tmp_l,con,wnd from weather_rcd_china
where con like '%雪%'
2025-05-21 人数最多的学生姓氏 
select left(name,1)as surname,count(1) as cnt
from students
group by
	surname
order by cnt desc
limit 5;
2025-05-21 多云天气天数 
select 
	city,
sum(if(con like '%多云%',1,0)) as cloudy_days,
concat(cast(sum(if(con like '%多云%',1,0))/count(1)*100 as decimal(4,2)),'%')as '占比'
from weather_rcd_china
where year(dt) = 2021
group by
	city
order by cloudy_days desc
2025-05-21 多云天气天数 
select 
	city,
sum(if(con like '%多云%',1,0)) as cloudy_days,
concat(cast(sum(if(con like '%多云%',1,0))/count(1) as decimal(4,2)),'%')as '占比'
from weather_rcd_china
where year(dt) = 2021
group by
	city
order by cloudy_days
2025-05-21 多云天气天数 
select 
	city,
sum(if(con like '%多云%',1,0)) as cloudy_days,
concat(cast(sum(if(con like '%多云%',1,0))/count(1) as decimal(4,2)),'%')as '占比'
from weather_rcd_china
where year(dt) = 2021
group by
	city
2025-05-21 德州扑克起手牌- 手对 
select * 
from hand_permutations
where left(card1,1) = left(card2,1)
2025-05-21 德州扑克起手牌- A花 
select * 
from hand_permutations
where (card1 like 'A%' or card2 like 'A%')and
	right(card1,1) = right(card2,1)
2025-05-21 德州扑克起手牌-最强起手牌KK+ 
select * 
from hand_permutations 
where card1 rlike '(K|A)' and card2 rlike '(K|A)'
2025-05-21 字符串与通配符(2)好多关键词做规则,可以使用rlike 
select 
	case
when mch_nm like '%按摩保健休闲%' then '按摩保健休闲'
whenlower(mch_nm) rlike '.*(按摩|保健|休闲|养生|spa|会所).*' then '按摩、保健、休闲、养生、SPA、会所'
end as reg_rules,
count(distinct mch_nm)
from cmb_usr_trx_rcd
where
lower(mch_nm) rlike '.*(按摩|保健|休闲|养生|spa|会所).*' or 
mch_nm like '%按摩保健休闲%'
group by
	reg_rules
2025-05-21 字符串与通配符(1)名称里面有特服,可以使用通配符 
select count(distinct mch_nm) as mch_cnt
from cmb_usr_trx_rcd
where mch_nm like '%按摩保健休闲%'
2025-05-21 字符串与通配符(1)名称里面有特服,可以使用通配符 
select count(*)
from cmb_usr_trx_rcd
where mch_nm like '%按摩保健休闲%'
2025-05-21 字符串与通配符(1)名称里面有特服,可以使用通配符 
select
	count(distinct if(mch_nm like '%按摩保健休闲%',mch_nm,null)) as mch_cnt
from cmb_usr_trx_rcd
2025-05-19 用户听歌习惯的时间分布 
select user_id,date_format(start_time,'%W')as day_of_week ,count(1)
from listen_rcd
group by user_id ,day_of_week
order by user_id,count(1) desc
2025-05-19 渣男腰子可真行,端午中秋干不停 
select * from cmb_usr_trx_rcd
where
	usr_id = '5201314520' and
	(date(trx_time) between '2024-06-08' and '2024-06-10' or date(trx_time) between '2024-09-15' and '2024-09-17')
order by trx_time