这是我的代码,但所有值都是相同的。
SELECT
department.name,
AVG(salary) AS Average,
SUM(salary) AS Total
FROM
department, employee
GROUP BY
department.name
答案 0 :(得分:0)
您错过了表格之间的加入条件,因此您将它们组合为完整的跨产品,而不是将员工仅链接到他们的部门。
SELECT d.name, AVG(e.salary), SUM(e.salary)
FROM department AS d
JOIN employee AS e ON d.dpt_no = e.dpt_no
GROUP BY d.name