标签:练习题 no employees dept emp SQL date NULL
本题来自于牛客网,答案为自己训练出,仅供参考。牛客网链接:https://www.nowcoder.com/activity/oj
题目一:
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,
PRIMARY KEY (`emp_no`));
测试数据集:
INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26');
INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21');
INSERT INTO employees VALUES(10003,'1959-12-03','Parto','Bamford','M','1986-08-28');
INSERT INTO employees VALUES(10004,'1954-05-01','Chirstian','Koblick','M','1986-12-01');
INSERT INTO employees VALUES(10005,'1955-01-21','Kyoichi','Maliniak','M','1989-09-12');
INSERT INTO employees VALUES(10006,'1953-04-20','Anneke','Preusig','F','1989-06-02');
INSERT INTO employees VALUES(10007,'1957-05-23','Tzvetan','Zielinski','F','1989-02-10');
INSERT INTO employees VALUES(10008,'1958-02-19','Saniya','Kalloufi','M','1994-09-15');
INSERT INTO employees VALUES(10009,'1952-04-19','Sumant','Peac','F','1985-02-18');
INSERT INTO employees VALUES(10010,'1963-06-01','Duangkaew','Piveteau','F','1989-08-24');
INSERT INTO employees VALUES(10011,'1953-11-07','Mary','Sluis','F','1990-01-22');
答案:
select * from employees order by hire_date desc limit 1
对应输出应该为:
10008|1958-02-19|Saniya|Kalloufi|M|1994-09-15
题目二:
查找入职员工时间排名倒数第三的员工所有信息(上表)
select * from employees order by hire_date desc limit 2,1
题目三:
查找各个部门当前(to_date='9999-01-01')领导当前薪水详情以及其对应部门编号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`));
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`));
答案:
select salaries.*, dept_no
from salaries left join dept_manager on salaries.emp_no = dept_manager.emp_no
where salaries.to_date='9999-01-01' AND
dept_manager.to_date='9999-01-01'
题目四:
查找所有已经分配部门的员工的last_name和first_name
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 `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`));
答案:
select last_name, first_name, dept_no
from dept_emp left join employees on dept_emp.emp_no = employees.emp_no
题目五:
查找所有员工的last_name和first_name以及对应部门编号dept_no,也包括展示没有分配具体部门的员工(上表)
答案:
select last_name, first_name, dept_no
from employees left join dept_emp on dept_emp.emp_no = employees.emp_no
题目六:
查找所有员工入职时候的薪水情况,给出emp_no以及salary, 并按照emp_no进行逆序
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`));
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`));
答案:
select employees.emp_no, salary
from employees left join salaries on employees.emp_no = salaries.emp_no
where employees.hire_date = salaries.from_date
order by employees.emp_no desc
题目七:
查找薪水涨幅超过15次的员工号emp_no以及其对应的涨幅次数t
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`));
答案:
select emp_no, count(salary) t from salaries
group by emp_no having t > 15
题目八:
标签:练习题,no,employees,dept,emp,SQL,date,NULL 来源: https://blog.51cto.com/u_15228753/2819773
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