Question

我正在使用ORACLE，并且此查询是在HR模式中完成的。我的问题是此查询并不完美，我迷路了，不知道如何继续。（请不要像您做作业那样粗鲁，因为我只会在尝试过东西时询问）。顺便说一句，这些陈述是西班牙语的，但我翻译了它们。

1.永久性媒体部门的高级职务和常任理事长。
英语：超过其所属部门的平均工资的雇员的雇员姓名和部门名称。

select e.first_name, e.last_name, d.department_name
    from employees e
    INNER JOIN departments d ON e.department_id = d.department_id
    where salary > (select avg(salary)
                        from employees
                        where department_id = department_id);

Empleados que，en cada departamento，tienen el sueldomáximo。
英语：每个部门中薪水最高的员工

select e.employee_id
    from employees e
    INNER JOIN departments d ON e.department_id = d.department_id
    where salary = (select hire_date
                        from employees
                        where employee_id = 107);

Empleados que trabajen en departamentos en los que nadie cobre comisiones。
英语：在没有人收取佣金的部门工作的员工

select e.employee_id
    from employees e
    INNER JOIN departments d ON e.department_id = d.department_id
    where commission_pct = (select sum(commission_pct)
                                from employees
                                group by department_id
                                having sum(commission_pct) = 0);

Taragao的第3部门，tra saljarios的总称，comisiones y的总称努马罗－德特拉巴贾多雷斯。 Mostrar el nombre del departamento y las 3 candidades。
英语：对于每个有3个或更多工人的部门，请计算总工资，佣金总额和工人人数。显示部门名称和3个数量

select sum(e.salary), sum(e.commission_pct), count(e.employee_id), d.department_name
    from employees e
    INNER JOIN departments d ON e.department_id = d.department_id
    where count(e.employee_id) >= 3;

Answer 1

在大多数情况下，您都非常亲密。

从1开始：应该始终使用表别名。如果不这样做，结果将是不可预测的，并且很可能是错误的。应该是：

select e.first_name, e.last_name, d.department_name
    from employees e
    INNER JOIN departments d ON e.department_id = d.department_id
    where e.salary > (select avg(e1.salary)
                        from employees e1
                        where e1.department_id = d.department_id);  --> aliases, especially here!

截至5：您的查询没有意义；怎么样？

select d.department_name, e.employee_id
from employees e INNER JOIN departments d ON e.department_id = d.department_id
where e.salary = (select max(e1.salary) 
                  from employees e1
                  where e1.department_id = d.department_id);

截至8：

select e.employee_id, d.department_name
    from employees e
    INNER JOIN departments d ON e.department_id = d.department_id
where e.department_id in (select e1.department_id
                          from employees e1
                          group by e1.department_id
                          having sum(commission_pct) is null;

截至10：

select sum(e.salary), sum(e.commission_pct), count(e.employee_id), d.department_name
    from employees e
    INNER JOIN departments d ON e.department_id = d.department_id
    group by d.department_name
    having count(*) >= 3;

我不知道如何纠正这个SQL练习

1 个答案: