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题解 | #创建一个actor_name表#

# create table if not exists actor_name( # first_name varchar(45) not null, # last_name varchar(45) not null # ) select first_name, last_name ...

2023-06-12

0 188

题解

# 子查询排序limit取第二id# select s.emp_no, salary, last_name, first_name# from employees e, salaries s# where salary = (# select salary# from salarie...

2023-06-02

0 209

题解

select f.film_id, title from film f left join film_category fc on f.film_id = fc.film_id where fc.category_id is null;

2023-06-02

0 189

题解 | #汇总各个部门当前员工的title类型的分配数目#

# 方法一，多表直接连接查询 # select de.dept_no, dept_name, title, count(t.emp_no) count # from departments d, dept_emp de, titles t # where d.dept_no = de.dept_no...

2023-06-02

0 223

题解 | #获取所有非manager员工当前的薪水情况#

# select de.dept_no, s.emp_no, s.salary # from salaries s, dept_emp de, dept_manager dm # where s.emp_no = de.emp_no and de.dept_no = dm.dept_no and d...

2023-06-02

1 207

题解

# SELECT e.last_name, e.first_name, # CASE # WHEN de.dept_no IS NULL THEN NULL # ELSE d.dept_name # END AS dept_name # FROM e...

2023-05-29

0 183

题解

select emp_no, salaryfrom salarieswhere salary = (select salary from salaries group by salary order by salary desc limit 1,1);# SELECT emp_no, salary#...

2023-05-29

0 222

题解 | #连续子数组最大和#

#include <climits> #include <iostream> #include <vector> using namespace std; int main() { int buf, n; cin >> n; ...

2023-04-21

0 295