sql练习题

Posted 2022-04-19 foremostxl

1、查找最晚入职员工的所有信息

CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,  只有年月日
PRIMARY KEY (`emp_no`));

所以同一入职的或许有很多人，order by files asc limit 1 是错误的
最晚入职的时间是最大的

select * from employees where hire_date = (select max(hire_date) from employees );
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mysql的日期类型分析

create table student(
    id int,
    name char(6),
    born_year year,
    birth_date date,
    class_time time,
    reg_time datetime
);

insert into student values
(1,‘egon‘,now(),now(),now(),now());

insert into student values
(2,‘alex‘,"1997","1997-12-12","12:12:12","2017-12-12 12:12:12");

2、查找入职员工时间排名倒数第三的员工所有信息

CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));

select * from employees where hire_date = (select hire_date from employees order by hire_date desc limit 2,1 );

注意limit 后面的 (n,m) n指索引 m是步长