有一个薪水表salaries简况如下: emp_no salary from_date to_date 10001 88958 2002-06-22 9999-01-01 10002 72527 2001-08-02 9999-01-01 10003 43311 2001-12-01 9999-01-01 请你获取薪水第二多的员工的emp_no以及其对应的薪水salary, 若有多个员工的薪水为第二多的薪水,则将对应的员工的emp_no和salary全部输出,并按emp_no升序排序。 emp_no salary 10002 72527
示例1

输入

drop table if exists  `salaries` ; 
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
INSERT INTO salaries VALUES(10001,88958,'2002-06-22','9999-01-01');
INSERT INTO salaries VALUES(10002,72527,'2001-08-02','9999-01-01');
INSERT INTO salaries VALUES(10003,43311,'2001-12-01','9999-01-01');

输出

10002|72527
加载中...