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题解 | #实习广场投递简历分析(二)#

select job, DATE_FORMAT(date, '%Y-%m') as mon, SUM(num) as cnt from resume_info where date between '2025-01-01' and '2025-12-31' group by job, mon or...

2023-09-04

0 246

题解 | #统计各个部门的工资记录数#

select d.dept_no, c.dept_name, count(d.emp_no) as sum from ( select b.dept_no, a.emp_no from salaries as a left join dept_emp as b on ...

2023-09-02

0 238

题解 | #牛客的课程订单分析(六)#

# 先找到满足条件的记录 select a.id, a.is_group_buy, b.name from ( select id, user_id, client_id, is_group_buy from order_info where user_id in ( ...

2023-09-02

0 294

题解 | #今天的刷题量(一)#

with t1 as ( select b.name, a.subject_id, create_time from submission as a left join subject as b on a.subject_id = b.id ) select name...

2023-09-02

0 303

题解 | 获取员工其当前的薪水比其manager当前薪水还高

# 先找到每个员工当前薪水 with t1 as ( select a.emp_no, a.dept_no, b.salary from dept_emp as a left join salaries as b on a.emp_no = b.emp_no ...

2023-09-02

0 354

题解 | #牛客的课程订单分析(五)#

# 先找到满足条件得记录 with t1 as ( select user_id, date from order_info where date > '2025-10-15' and status = 'completed' and product_...

2023-09-01

1 245

题解 | RIGHT提取右边指定长度字符串

select first_name from ( select first_name, RIGHT(first_name, 2) as last_two_string from employees order by last_two_string ) as a;

2023-08-31

0 350

题解 | 原字符串长度-逗号替换为空字符串后的字符串长度

select id, LENGTH(string)-LENGTH(REPLACE(string,',','')) as cnt from strings;

2023-08-31

0 294

题解 | #分页查询employees表，

# 也就是需要返回6到10行的数据 select * from employees limit 5, 5;

2023-08-31

0 325

题解 | #考试分数(二)#

# 先找到每一个工作的平均分 with t1 as ( select job, AVG(score) as avg_score from grade group by job ) select id, job, score from ( select a.*, t1....

2023-08-30

0 326