Question

我有两张桌子：

Departament

id_dep, name_dep

员工

id_dep, salary

我想获得所有工资都低于7000的部门的名称。我必须创建一个SELECT语句。我是SQL的初学者，如果我会浪费你的时间，我很抱歉，但我需要帮助。

我有一个解决方案，但我不知道是否好：

SELECT name_dep
FROM departments
WHERE employee.salary < 7000
MINUS
SELECT name_dep
FROM departments
WHERE employee.salary >= 7000

Answer 1

您可以使用NOT EXISTS执行此操作：

SELECT id_dep , name_dep
FROM department AS d
WHERE NOT EXISTS (SELECT 1 
                  FROM employee AS e
                  WHERE e.id_dep = d.id_dep AND e.salary >=  7000)

Answer 2

将JOINs添加到原始MINUS次尝试中：

SELECT d.name_dep
FROM departments d
  JOIN employee e ON d.id_dep = e.id_dep
WHERE e.salary < 7000
MINUS
SELECT d.name_dep
FROM departments d
  JOIN employee e ON d.id_dep = e.id_dep
WHERE e.salary >= 7000

Answer 3

您可以使用NOT EXISTS或EXISTS，针对您的问题尝试此代码块。

NOT EXISTS

select d.name_dep
  from department as d
 where not exists (select 1
          from employee as e
         where e.id_dep = d.id_dep
           and e.salary >= 7000)
 group by d.dep_name

<强> EXISTS

select d.name_dep
  from department as d
 where exists (select 1
          from employee as e
         where e.id_dep = d.id_dep
           and e.salary < 7000)
 group by d.dep_name

Answer 4

您可以按部门分组并检查组中的最高工资：

select d.id_dep, d.name_dep
from department d
join employee e on d.id_dep = e.id_dep
group by d.id_dep, d.name_dep
having max(e.salary) < 700

SQL SELECT（初学者）

4 个答案: