查找总工资大于所有部门总工资平均值的所有部门答案

【问题标题】：Finding all departments where the total salary is greater than the average of the total salary at all departments查找总工资大于所有部门总工资平均值的所有部门
【发布时间】：2016-07-07 15:36:24
【问题描述】：

我发现以前问过类似类型的问题。但这些问题并不完美，我想要的。所以，我必须在这里问。

我有五个表如下所示...

department(dept_name,building,budget);
primary key (dept_name)

course(course_id,title,dept_name,credits);
primary key (course_id),
foreign key (dept_name) references department

instructor(id,name,dept_name,salary);
primary key (ID),
foreign key (dept_name) references department

section(course_id,sec_id,semester,year,building,room_number,time_slot_id);
primary key (course_id, sec_id, semester, year),
foreign key (course_id) references course

teaches(id,course_id,sec_id,semester,year);
primary key (ID, course_id, sec_id, semester, year),
foreign key (course_id, sec_id, semester, year) references section,
foreign key (ID) references instructor

现在我正在查找总薪水大于总薪水平均值的所有部门使用此查询的所有部门。

with dept_total (dept_name, value) as
(select dept_name, sum(salary)
from instructor
group by dept_name),
dept_total_avg(value) as
(select avg(value)
from dept_total)
select dept_name
from dept_total, dept_total_avg
where dept_total.value >= dept_total_avg.value;

不使用提供相同结果的 with 子句的等效查询是什么？

【问题讨论】：

标签： sql database oracle11g relational-database

【解决方案1】：

使用 HAVING 子句将允许这样做，以下 SQL 未经测试但应该可以工作...

SELECT d.dept_name,
    SUM(salary) AS DeptTotalSalary,
    SUM(salary) / COUNT(*) AS DeptAverageSalary
FROM department d
JOIN instructor i ON i.dept_name d.dept_name
GROUP BY d.dept_name
HAVING DeptAverageSalary > DeptTotalSalary

【讨论】：

【解决方案2】：

我没有要测试的更改，但这样的东西应该可以工作。

SELECT dept_name FROM instructor 
GROUP BY dept_name
HAVING SUM(salary)>=
(SELECT AVG(dept_sum.SumSalary) AS AvgSalary FROM 
    (
        SELECT dept_name, sum(salary) as SumSalary FROM instructor
        GROUP BY dept_name
    ) as dept_sum
)

【讨论】：

运行此查询后显示ERROR at line 8: ORA-00907: missing right parenthesis。
虽然我没有发现这里缺少任何括号。

【解决方案3】：

最后，我得到了正确的查询。所以，不使用 with 子句的等效查询将是这个。

select dept_name
from (
    select dept_name, sum(salary) dept_total
    from instructor
    group by dept_name
), (
    select avg(dept_total) dept_total_avg
    from (
        select dept_name, sum(salary) dept_total
        from instructor
        group by dept_name
    )
) where dept_total >= dept_total_avg;

【讨论】：