Oracle数据库emp查询job_id的员工最高工资和最低工资的差距 （DIFFERENCE）？

问题:oracle经典emp表,若要找出每种JOB中工资最高的员工的记录,该如何.

select empno,ename,emp.job,maxsal sal,deptnofrom emp join (select job,max(sal) maxsal from emp group by job) t on emp.sal=t.maxsal;

oracle数据库系统表emp的查询:求出工资比我当前员工高出500以外的人.

SELECT COUNT(*)FROM employees aWHERE a.salary-500>(SELECT salary FROM employees WHERE employees.employee_id=&a);---由于不知道你当前员工的ID是多少,所以用了变量a 工资比我当前员工高出500以外: a.salary-500>(.) 这个要注意

Oracle 查询工资最高和最低的工种,用sql语句表达出来,谢谢.

SELECT max(salay) '最高工资',min(salay) '最低工资' FROM salayTable GROUP BY salay

oracle的Emp和dept表下根据部门编号查询员工工资最低的然后.

update EMPLOYEES set salary=salary*1.1 where (department_id,salary) in(select department_id,min(salary) from EMPLOYEES group by department_id);

对oracle数据库系统表emp的查询:显示平均工资低于2000的部门号、平均.

select deptNo from emp group by deptNo having avg(salary)评论0 00

在oracle中 “查询并显示每个部门的最高工资、最低工资、并按.

你好!一句代码就叫搞定了:select deptno, max(sal) maxSal,min(sal) minSal from emp group by deptno order by deptno desc 如有疑问,请追问.

oracle数据库中怎么查询emp表中工资最大的三个人的全部信息?

表结构都没有,怎么写? SELECT empno,ename,sal,job,mgr,hiredate,deptno FROM (SELECT empno,ename,sal,job,mgr,hiredate,deptnoFROM emp ORDER BY sal DESC) t WHERE rownum <= 3

oracle要查询每个部门里工资大于该部门的平均工资的员工,要.

select employees.employee_id, employees.salary, employees.department_id, avg_tmp.avgsalary from employees left outer join (select department_id, avg(salary) as avgsalary from employees group by department_id) avg_tmp on employees.department_id = avg_tmp.department_id

查询各部门的部门编号及该部门中薪水等级最高的职位.这是ora.

消灭0回答 没有Oracle 没执行 你试试 大概就这个思路吧 select e.职位,d.部门编号 from dept d, emp e, salgrade s where e.员工编号=s.员工编号 and e.部门编号=d.部门编号 group by d.部门编号 having s.薪水=MAX(s.薪水)

oracle语句,查询最低薪金大于1400的工作的最低薪金;

select min(sal) from emp where sal>1400