这才是开发的壁垒，公司倒闭都没一人能看懂~

今天我们要聊一聊那种长得像“论文”的SQL查询。

要是你也曾被这种巨无霸SQL搞得头昏脑涨，那你肯定会对这篇文章有共鸣。

如图：是不是已经头昏脑胀了？

咱们就借着这张令人瞠目结舌的SQL截图，来细细剖析一下这背后的门道。

首先，看看这张图，我只能说“这哪里是SQL啊，这简直是SQL界的史诗级论文！” 😱

-- 原始SQL（略作简化）SELECT a.name, b.salary, c.department_name FROM employees aJOIN salaries b ON a.employee_id = b.employee_idJOIN departments c ON a.department_id = c.department_idWHERE a.status = 'ACTIVE'AND b.salary > (SELECT AVG(salary) FROM salaries WHERE department_id = c.department_id)ORDER BY b.salary DESC;

-- 创建临时表/视图CREATE VIEW avg_salaries ASSELECT department_id, AVG(salary) AS avg_salaryFROM salariesGROUP BY department_id;
-- 使用视图简化主查询SELECT a.name, b.salary, c.department_name FROM employees aJOIN salaries b ON a.employee_id = b.employee_idJOIN departments c ON a.department_id = c.department_idJOIN avg_salaries d ON c.department_id = d.department_idWHERE a.status = 'ACTIVE'AND b.salary > d.avg_salaryORDER BY b.salary DESC;

-- 筛选出所有在职员工SELECT a.name, b.salary, c.department_name FROM employees a-- 获取员工薪资信息JOIN salaries b ON a.employee_id = b.employee_id-- 获取员工所在部门信息JOIN departments c ON a.department_id = c.department_id-- 获取部门平均薪资JOIN avg_salaries d ON c.department_id = d.department_id-- 筛选薪资高于部门平均值的员工WHERE a.status = 'ACTIVE'AND b.salary > d.avg_salaryORDER BY b.salary DESC;

# 结语