coding的佩奇酱

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题解 | 统计所有课程参加培训人次

select round(sum(length(replace(coalesce(course,0),',','')) / 7),0) as staff_nums from cultivate_tb;

2026-02-06

0 23

题解 | 分析客户逾期情况

with tmp as (select t1.customer_id, t1.pay_ability, case when overdue_days is null then 0 else 1 end as num, count(1) over(partition by pay_ability) ...

2026-02-03

0 21

题解 | 最长连续登录天数

select user_id, max(maxrn) as max_consec_days from (select user_id, datediff ,count(datediff) as maxrn from (select fdate, user_id, fdate - rn as da...

2026-02-02

0 34

题解 | 用列表实现队列

queue = [1, 2, 3, 4, 5] i=0 n=int(input()) while i<=1: queue.pop(0) print(queue) i+=1 queue.append(n) print(queue)

2026-01-31

0 25

题解 | 每个月Top3的周杰伦歌曲

select * from (select month, row_number() over(partition by month order by play_pv desc,song_id asc) as ranking, song_name, play_pv from (with tmp as ...

2026-01-22

0 31

题解 | 基本数学函数

select id, value, case when value >= 0 then value when value < 0 then value *-1 end as absolute_value, ceil(value) as ceiling_value, ...

2026-01-14

0 25

题解 | 电话号码格式校验

select id, name, phone_number from contacts where (length(phone_number) = 10 or length(replace(phone_number,'-','')) = 10) and substring(phone_number,...

2026-01-14

0 33

题解 | #平均工资#

select avg(salary) from salaries where salary != (select max(salary) from salaries where to_date = '9999-01-01') and salary != (select min(salary) fr...

2023-08-31

0 259

题解 | #牛客的课程订单分析(五)#

select t3.user_id, t3.first_buy_date, tc.second_buy_date, t3.cnt from (select * from (select user_id, case when rn = 1 then date else null end as fir...

2023-08-29

0 390

题解 | #牛客的课程订单分析(六)#

select id,is_group_buy,name from (select id,is_group_buy,name,max(rn) over(partition by user_id) mx from ( select t1.id, t1.user_id, t1.is_group_buy...

2023-08-24

0 316