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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
2026-01-08 基于共同兴趣爱好的餐厅推荐（1）-我吃过啥  已解决
2026-01-07 基础标量子查询-带分组  已解决
2026-01-07 名字中字母e左起小于等于3位的歌手  已解决
2026-01-06 登录天数分布  已解决
2026-01-05 通勤、午休、临睡个时间段活跃人数分布  已解决
2025-12-26 滴滴面试真题（2）打车订单呼叫应答时间  已解决
2025-12-26 抖音面试真题（1）T+1日留存率  已解决
2025-12-25 绘制小时进入人数曲线  已解决
2025-12-23 一线城市历年平均气温  已解决
2025-12-22 分组与聚合函数（5）五花八门的项目，其实都有固定套路（2）  已解决
2025-12-22 查询播放量为0的歌手及其专辑  已解决
2025-12-21 小结-行转列，展开学生成绩（1）  已解决
2025-12-21 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
2025-12-25 绘制小时进入人数曲线  
这样写不就可以了么
SELECT 
	HOUR(enter_time) AS hour_entered,
	COUNT(usr_id) AS enter_count
FROM ks_live_t1
GROUP BY HOUR(enter_time)
ORDER BY HOUR(enter_time)
 
啥也没说

提交记录

提交日期 题目名称 提交代码
2026-01-08 抖音面试真题（1）T+1日留存率  
SELECT T_login AS login_date,
CONCAT(ROUND(AVG(T_1_login IS NOT NULL)*100,2),'%') AS T1_retention_rate
FROM (SELECT DISTINCT T.usr_id, 
 DATE(T.login_time) AS T_login, 
 DATE(T_1.login_time) AS T_1_login
 FROM user_login_log T
 LEFT JOIN 
 user_login_log T_1
 ON T.usr_id = T_1.usr_id AND DATEDIFF(T_1.login_time, T.login_time) = 1
 WHERE T.login_time > DATE_SUB(current_date(), INTERVAL 30 DAY))c
 GROUP BY T_login
 ORDER BY login_date
2026-01-08 抖音面试真题（1）T+1日留存率  
SELECT T_login AS login_date,
CONCAT(ROUND(AVG(T_1_login IS NOT NULL)*100,2),'%') AS T1_retention_rate
FROM (SELECT a.usr_id, 
 DATE(a.login_time) AS T_login, 
 DATE(b.login_time) AS T_1_login
 FROM user_login_log a
 LEFT JOIN 
 user_login_log b
 ON a.usr_id = b.usr_id AND DATEDIFF(b.login_time, a.login_time) = 1
 WHERE a.login_time > DATE_SUB(CURDATE(), INTERVAL 30 DAY))a
 GROUP BY T_login
2026-01-08 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 c<=0 and -b/(2*a) > 0 )
2026-01-08 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 c<=0 and -b/(2*a) > 0 )
2026-01-08 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 c<=0)
2026-01-08 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
2026-01-08 基于共同兴趣爱好的餐厅推荐（1）-我吃过啥  
select DISTINCT cust_uid,
 mch_nm 
from mt_trx_rcd1
WHERE cust_uid = 'MT10000'
2026-01-08 基于共同兴趣爱好的餐厅推荐（1）-我吃过啥  
select cust_uid,
mch_nm from mt_trx_rcd1
WHERE cust_uid = 'MT10000'
2026-01-08 经过3个象限的一元一次函数  
select 
    * 
from
    numbers_for_fun
where
    a = 0 and b<>0 and c<>0
order by 1
2026-01-08 经过至少两个象限的一元一次函数  
select 
    * 
from
    numbers_for_fun
where
    a = 0 and b<>0 and c<>0
2026-01-08 一元一次函数形成的三角形面积  
select * from numbers_for_fun 
where a = 0 and b != 0 and c != 0
and 0.5*abs(c*c/b)>=5
2026-01-08 一元一次函数形成的三角形面积  
select * from numbers_for_fun 
where a = 0 and b != 0 and c != 0
and (c*c/b<=-10) or (c*c/b>=10)
2026-01-08 一元一次函数形成的等腰三角形  
select * from numbers_for_fun 
where a = 0 and ((b = 1) or (b = -1)) and c != 0
2026-01-08 一元一次函数形成的等腰三角形  
select * from numbers_for_fun 
where a = 0 and ((b = 1) or (b = -1))
2026-01-08 一元一次函数形成的等腰三角形  
select * from numbers_for_fun 
where a = 0 and (b = 1) or (b = -1)
2026-01-08 HAVING子句中的子查询  
select mch_nm, 
 sum(trx_amt) as mch_total_amt
 from cmb_usr_trx_rcd
 group by mch_nm 
 having mch_total_amt > 10*(select avg(trx_amt) from cmb_usr_trx_rcd)
2026-01-08 NOT IN子查询  
select * from cmb_usr_trx_rcd 
where usr_id NOT IN (select DISTINCT usr_id from cmb_usr_trx_rcd WHERE mch_nm = '肯德基'
 )
2026-01-08 IN子查询  
SELECT * FROM cmb_usr_trx_rcd 
WHERE usr_id in (select DISTINCT usr_id from cmb_usr_trx_rcd where mch_nm = '肯德基')
2026-01-08 IN子查询  
select * from cmb_usr_trx_rcd where mch_nm = '肯德基'
2026-01-08 WHERE子查询 + 比较运算符  
select * from cmb_usr_trx_rcd 
where trx_amt > (select avg(trx_amt) from cmb_usr_trx_rcd)