select tab1.dept_no,tab1.emp_no,tab2.salary from ( select * from dept_emp where emp_no not in ( select distinct emp_no from dept_manager ) ) as tab1 left join salaries as tab2 on tab1.emp_no=tab2.emp_no; 分析想要结果中的几个字段分别来自哪几张表,确定数据来源。找出所有manager的员工号。提取表dept_emp中不属于...