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题解 | 查找最晚入职员工的所有信息

select * from employees order by hire_date desc limit 1 #代表取1行数据 offset 0 #从第几+1行开始取 limit 1 限制取几行offset 0 从n+1行开始取即从第一行开始取

2025-02-21

0 34

题解 | #判断其是否有过购买记录#

select customer_id , if (latest_place_order_date is null, 0,1) as if_placed_order from customers_info

2024-10-09

0 65

题解 | #查询职位发布时间在2021年后

select job_id , boss_id , company_id from job_info where year(post_time)>='2021' or job_city like "上海%" order by job_ci...

2024-10-09

0 97

和company_id，结果不去重#

二者合并输出不去重..union all 俩个表上下合并

2024-10-09

0 88

题解 | #计算刷题总人数#

select count(distinct user_id) as cnt from questions_pass_record_detail 注意：要对用户id先去重再计数

2024-10-09

0 67

题解 | #查询城市为北京的职位投递记录#

select * from deliver_record_detail where job_city like "北京%" 通配符的使用需要与like相结合，%表示零个或多个字符，_表示一个字符。

2024-10-09

0 81

\

select user_id from deliver_record_detail where coalesce(max_salary,0) - coalesce(min_salary,0) > 2 注意：处理空值数据coalesce（）标准SQL，infull()方言MYSQL

2024-10-09

0 78

题解 | #考试分数(五)#

select t1.id,t1.job,t1.score,t1.rk from (select id,job,score,row_number() over (partition by job order by score desc) rk from grade ) t1 inner join (s...

2024-09-03

0 101

题解 | #考试分数(四)#

with t as (select job,count(score) as num from grade group by job) select job , ceil(num/2) as start , if(num%2=0,ceil(num/2)+1,ceil(num/2)...

2024-09-03

0 101

题解 | #考试分数(三)#

select g1.id, a.name, g1.score from grade as g1 inner join language as a on g1.language_id=a.id where ( select count(distinct g2.score) from g...

2024-09-02

0 79