mysql数据库面试题(3)
题目描述
11.获取所有员工当前的manager,如果当前的manager是自己的话结果不显示,当前表示to_date=‘9999-01-01‘。结果第一列给出当前员工的emp_no,第二列给出其manager对应的manager_no。
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `dept_manager` (
`dept_no` char(4) NOT NULL,
`emp_no` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
输入
select d.emp_no, m.emp_no as manager_no
from dept_emp d inner join dept_manager m
on d.dept_no = m.dept_no
where m.to_date = ‘9999-01-01‘ and d.to_date = ‘9999-01-01‘ and d.emp_no <> m.emp_no
输出
emp_no | manager_no |
10001 | 10002 |
10003 | 10004 |
10009 | 10010 |
12.获取所有部门中当前员工薪水最高的相关信息,给出dept_no, emp_no以及其对应的salary
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
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`));
输入
SELECT d.dept_no,d.emp_no,MAX(s.salary) AS salary
FROM salaries s inner join dept_emp d
ON d.emp_no=s.emp_no
WHERE d.to_date=‘9999-01-01‘ AND s.to_date=‘9999-01-01‘
GROUP BY d.dept_no
输出
dept_no | emp_no | salary |
d001 | 10001 | 88958 |
d002 | 10006 | 43311 |
d003 | 10005 | 94692 |
d004 | 10004 | 74057 |
d005 | 10007 | 88070 |
d006 | 10009 | 95409 |
13.从titles表获取按照title进行分组,每组个数大于等于2,给出title以及对应的数目t。
CREATE TABLE IF NOT EXISTS "titles" (
`emp_no` int(11) NOT NULL,
`title` varchar(50) NOT NULL,
`from_date` date NOT NULL,
`to_date` date DEFAULT NULL);
输入
select title,count(title) as t from titles
group by title having t>=2;
输出
itle | t |
Assistant Engineer | 2 |
Engineer | 4 |
省略 | 省略 |
Staff | 3 |
14.从titles表获取按照title进行分组,每组个数大于等于2,给出title以及对应的数目t。注意对于重复的emp_no进行忽略。
CREATE TABLE IF NOT EXISTS `titles` (
`emp_no` int(11) NOT NULL,
`title` varchar(50) NOT NULL,
`from_date` date NOT NULL,
`to_date` date DEFAULT NULL);
输入
SELECT title,COUNT(DISTINCT emp_no) AS t
FROM titles
GROUP BY title
HAVING t >= 2;
输出
title | t |
Assistant Engineer | 2 |
Engineer | 3 |
省略 | 省略 |
Staff | 3 |
15.查找employees表所有emp_no为奇数,且last_name不为Mary的员工信息,并按照hire_date逆序排列
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`));
输入
select * from employees
where emp_no%2!=0 and last_name!=‘Mary‘
order by hire_date desc;
输出
emp_no | birth_date | first_name | last_name | gender | hire_date |
10011 | 1953-11-07 | Mary | Sluis | F | 1990-01-22 |
10005 | 1955-01-21 | Kyoichi | Maliniak | M | 1989-09-12 |
10007 | 1957-05-23 | Tzvetan | Zielinski | F | 1989-02-10 |
10003 | 1959-12-03 | Parto | Bamford | M | 1986-08-28 |
10001 | 1953-09-02 | Georgi | Facello | M | 1986-06-26 |
10009 | 1952-04-19 | Sumant | Peac | F | 1985-02-18 |