mysql学习打卡day22
今日成果：
select * from employees
where salary > (select avg(salary) from employees);
-- 查询超过平均工资的员工
select * from clients
where client_id not in (select distinct client_id from invoices);
-- 查询没有发票的用户 感谢各位读者查阅，欢迎各位👍点赞✍评论⭐收藏+关注！
select * from employees
where salary > (select avg(salary) from employees);
-- 查询超过平均工资的员工
select * from clients
where client_id not in (select distinct client_id from invoices);
-- 查询没有发票的用户 感谢各位读者查阅，欢迎各位👍点赞✍评论⭐收藏+关注！