Question

我之前发现了类似的问题类型。但那些并不完美，我想要的。所以，我在这里问过。

我有五张桌子如下......

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

现在我发现所有部门的总薪水都高于总薪水的平均值在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;

如果不使用提供相同结果的with子句，那么等效查询是什么？

Answer 1

使用HAVING子句将允许此操作，以下SQL未经测试但应该有效...

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

Answer 2

我没有改变测试，但这样的事情应该有效。

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
)

Answer 3

最后，我得到了正确的查询。因此，不使用with子句的等效查询就是这个。

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;

查找总薪资大于所有部门总薪资平均值的所有部门

3 个答案: