<MySQL数据库>牛客网的SQL实战
1.查找最晚入职员工的所有信息（减轻入门难度：目前所有的数据里员工入职的日期都不是同一天）
（1）创建表
CREATE TABLE `employees` ( `emp_no` int(11) NOT NULL, -- '员工编号' `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`));
（2）插入数据
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');
（3）查询最晚入职员工的所有信息
方法一
SELECT * FROM employees WHERE hire_date = (SELECT max(hire_date) FROM employees);
运行效果：
方法二
select * from employees order by hire_date desc limit 1;
运行效果：
2.查找入职员工时间排名倒数第三的员工所有信息（减轻入门难度：目前所有的数据里员工入职的日期都不是同一天）
LIMIT m,n : 表示从第m+1条开始，取n条数据；
LIMIT n ： 表示从第0条开始，取n条数据，是limit(0,n)的缩写
SELECT * FROM employees WHERE hire_date = (SELECT distinct hire_date FROM employees order by hire_date desc LIMIT 2,1);
3.查找各个部门当前(dept_manager.to_date='9999-01-01')领导当前(salaries.to_date='9999-01-01')薪水详情以及其对应部门编号dept_no
(注:输出结果以salaries.emp_no升序排序，并且请注意输出结果里面dept_no列是最后一列)
(1)创建表
CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL, -- '员工编号', `salary` int(11) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`from_date`)); CREATE TABLE `dept_manager` ( `dept_no` char(4) NOT NULL, -- '部门编号' `emp_no` int(11) NOT NULL, -- '员工编号' `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`dept_no`));
(3)查询结果
select a.*, b.dept_no from salaries as a left join dept_manager as b on a.emp_no=b.emp_no where a.to_date='9999-01-01' and b.to_date='9999-01-01' order by a.emp_no asc;
4.查找所有已经分配部门的员工的last_name和first_name以及dept_no(请注意输出描述里各个列的前后顺序)
(1)创建表
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`));
(3)查询结果
select a.last_name,a.first_name,b.dept_no from employees a left join dept_emp b where a.emp_no=b.emp_no;
5.查找所有员工的last_name和first_name以及对应部门编号dept_no，也包括暂时没有分配具体部门的员工(请注意输出描述里各个列的前后顺序)
(1)创建表格
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`));
（2）查询结果
select a.last_name,a.first_name,b.dept_no from employees a left join dept_emp b on a.emp_no=b.emp_no;
6.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(请注意，一个员工可能有多次涨薪的情况)
（1）创建表格
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`));
（2）查询结果
select a.emp_no,b.salary from employees a left join salaries b on a.emp_no=b.emp_no where a.hire_date=b.from_date order by a.emp_no desc;
7.查找薪水变动超过15次的员工号emp_no以及其对应的变动次数t
(1)创建表
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`));
(2)查询结果
select emp_no,count(salary) t from salaries group by emp_no having count(salary)>15;
8.找出所有员工当前(to_date='9999-01-01')具体的薪水salary情况，对于相同的薪水只显示一次,并按照逆序显示
（1）创建表
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`));
（2）查询结果
select distinct salary from salaries where to_date='9999-01-01' order by salary desc;
9.获取所有部门当前(dept_manager.to_date='9999-01-01')manager的当前(salaries.to_date='9999-01-01')薪水情况，给出dept_no, emp_no以及salary(请注意，同一个人可能有多条薪水情况记录)
(1)创建表
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`));
（2）查询结果
select a.dept_no,a.emp_no,b.salary from dept_manager a left join salaries b on a.emp_no=b.emp_no where a.to_date='9999-01-01' and b.to_date='9999-01-01';
10.获取所有非manager的员工emp_no
（1）创建表
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 `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`));
（2）插入数据
INSERT INTO dept_manager VALUES('d001',10002,'1996-08-03','9999-01-01');
INSERT INTO dept_manager VALUES('d002',10006,'1990-08-05','9999-01-01');
INSERT INTO dept_manager VALUES('d003',10005,'1989-09-12','9999-01-01');
INSERT INTO dept_manager VALUES('d004',10004,'1986-12-01','9999-01-01');
INSERT INTO dept_manager VALUES('d005',10010,'1996-11-24','2000-06-26');
INSERT INTO dept_manager VALUES('d006',10010,'2000-06-26','9999-01-01');
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');
（3）查询结果
select emp_no from employees where emp_no not in (select emp_no from dept_manager);
11.获取所有员工当前的(dept_manager.to_date='9999-01-01')manager，如果员工是manager的话不显示(也就是如果当前的manager是自己的话结果不显示)。输出结果第一列给出当前员工的emp_no,第二列给出其manager对应的emp_no。
（1）创建表格
CREATE TABLE `dept_emp` ( `emp_no` int(11) NOT NULL, -- '所有的员工编号' `dept_no` char(4) NOT NULL, -- '部门编号' `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, -- '部门编号' `emp_no` int(11) NOT NULL, -- '经理编号' `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`dept_no`));
(2)插入数据
INSERT INTO dept_emp VALUES(10001,'d001','1986-06-26','9999-01-01');
INSERT INTO dept_emp VALUES(10002,'d001','1996-08-03','9999-01-01');
INSERT INTO dept_emp VALUES(10003,'d004','1995-12-03','9999-01-01');
INSERT INTO dept_emp VALUES(10004,'d004','1986-12-01','9999-01-01');
INSERT INTO dept_emp VALUES(10005,'d003','1989-09-12','9999-01-01');
INSERT INTO dept_emp VALUES(10006,'d002','1990-08-05','9999-01-01');
INSERT INTO dept_emp VALUES(10007,'d005','1989-02-10','9999-01-01');
INSERT INTO dept_emp VALUES(10008,'d005','1998-03-11','2000-07-31');
INSERT INTO dept_emp VALUES(10009,'d006','1985-02-18','9999-01-01');
INSERT INTO dept_emp VALUES(10010,'d005','1996-11-24','2000-06-26');
INSERT INTO dept_emp VALUES(10010,'d006','2000-06-26','9999-01-01');
INSERT INTO dept_manager VALUES('d001',10002,'1996-08-03','9999-01-01');
INSERT INTO dept_manager VALUES('d002',10006,'1990-08-05','9999-01-01');
INSERT INTO dept_manager VALUES('d003',10005,'1989-09-12','9999-01-01');
INSERT INTO dept_manager VALUES('d004',10004,'1986-12-01','9999-01-01');
INSERT INTO dept_manager VALUES('d005',10010,'1996-11-24','2000-06-26');
INSERT INTO dept_manager VALUES('d006',10010,'2000-06-26','9999-01-01');
(3)查询结果
select a.emp_no,b.emp_no from dept_emp a left join dept_manager b on a.dept_no=b.dept_no where b.to_date="9999-01-01" and a.emp_no!=b.emp_no;
12.获取所有部门中当前(dept_emp.to_date = '9999-01-01')员工当前(salaries.to_date='9999-01-01')薪水最高的相关信息，给出dept_no, emp_no以及其对应的salary，按照部门升序排列。
(1)创建表格
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`));
（2）插入数据
INSERT INTO dept_emp VALUES(10001,'d001','1986-06-26','9999-01-01'); INSERT INTO dept_emp VALUES(10002,'d001','1996-08-03','9999-01-01'); INSERT INTO dept_emp VALUES(10003,'d001','1996-08-03','1997-08-03'); INSERT INTO salaries VALUES(10001,90000,'1986-06-26','1987-06-26'); INSERT INTO salaries VALUES(10001,88958,'2002-06-22','9999-01-01'); INSERT INTO salaries VALUES(10002,72527,'1996-08-03','1997-08-03'); INSERT INTO salaries VALUES(10002,72527,'2000-08-02','2001-08-02'); INSERT INTO salaries VALUES(10002,72527,'2001-08-02','9999-01-01'); INSERT INTO salaries VALUES(10003,90000,'1996-08-03','1997-08-03');
（3）查询结果
select a.dept_no,a.emp_no,MAX(b.salary) from dept_emp a inner join salaries b on a.emp_no=b.emp_no where a.to_date='9999-01-01' and b.to_date='9999-01-01' GROUP by a.dept_no
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。
（1）创建表格
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);
（2）插入数据
INSERT INTO titles VALUES(10001,'Senior Engineer','1986-06-26','9999-01-01'); INSERT INTO titles VALUES(10002,'Staff','1996-08-03','9999-01-01'); INSERT INTO titles VALUES(10003,'Senior Engineer','1995-12-03','9999-01-01'); INSERT INTO titles VALUES(10004,'Engineer','1986-12-01','1995-12-01'); INSERT INTO titles VALUES(10004,'Senior Engineer','1995-12-01','9999-01-01'); INSERT INTO titles VALUES(10005,'Senior Staff','1996-09-12','9999-01-01'); INSERT INTO titles VALUES(10005,'Staff','1989-09-12','1996-09-12'); INSERT INTO titles VALUES(10006,'Senior Engineer','1990-08-05','9999-01-01'); INSERT INTO titles VALUES(10007,'Senior Staff','1996-02-11','9999-01-01'); INSERT INTO titles VALUES(10007,'Staff','1989-02-10','1996-02-11'); INSERT INTO titles VALUES(10008,'Assistant Engineer','1998-03-11','2000-07-31'); INSERT INTO titles VALUES(10009,'Assistant Engineer','1985-02-18','1990-02-18'); INSERT INTO titles VALUES(10009,'Engineer','1990-02-18','1995-02-18'); INSERT INTO titles VALUES(10009,'Senior Engineer','1995-02-18','9999-01-01'); INSERT INTO titles VALUES(10010,'Engineer','1996-11-24','9999-01-01'); INSERT INTO titles VALUES(10010,'Engineer','1996-11-24','9999-01-01');
（3）查询结果
select title,count(title) as t from titles group by title having count(title)>=2;
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。对于重复的emp_no进行忽略(即emp_no重复的title不计算，title对应的数目t不增加)。
（1）创建表
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);
（2）插入数据
INSERT INTO titles VALUES(10001,'Senior Engineer','1986-06-26','9999-01-01'); INSERT INTO titles VALUES(10002,'Staff','1996-08-03','9999-01-01'); INSERT INTO titles VALUES(10003,'Senior Engineer','1995-12-03','9999-01-01'); INSERT INTO titles VALUES(10004,'Engineer','1986-12-01','1995-12-01'); INSERT INTO titles VALUES(10004,'Senior Engineer','1995-12-01','9999-01-01'); INSERT INTO titles VALUES(10005,'Senior Staff','1996-09-12','9999-01-01'); INSERT INTO titles VALUES(10005,'Staff','1989-09-12','1996-09-12'); INSERT INTO titles VALUES(10006,'Senior Engineer','1990-08-05','9999-01-01'); INSERT INTO titles VALUES(10007,'Senior Staff','1996-02-11','9999-01-01'); INSERT INTO titles VALUES(10007,'Staff','1989-02-10','1996-02-11'); INSERT INTO titles VALUES(10008,'Assistant Engineer','1998-03-11','2000-07-31'); INSERT INTO titles VALUES(10009,'Assistant Engineer','1985-02-18','1990-02-18'); INSERT INTO titles VALUES(10009,'Engineer','1990-02-18','1995-02-18'); INSERT INTO titles VALUES(10009,'Senior Engineer','1995-02-18','9999-01-01'); INSERT INTO titles VALUES(10010,'Engineer','1996-11-24','9999-01-01'); INSERT INTO titles VALUES(10010,'Engineer','1996-11-24','9999-01-01');
（3）查询结果
select title, count(distinct emp_no) from titles group by title having count(title)>=2;
15.查找employees表所有emp_no为奇数，且last_name不为Mary(注意大小写)的员工信息，并按照hire_date逆序排列(题目不能使用mod函数)
（1）创建表格
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`));
（2）插入数据
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');
（3）查询结果
select * from employees where emp_no%2=1 and last_name!='Mary' order by hire_date desc;
16.统计出当前(titles.to_date='9999-01-01')各个title类型对应的员工当前(salaries.to_date='9999-01-01')薪水对应的平均工资。结果给出title以及平均工资avg。
（1）创建表格
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`)); 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);
（2）插入数据
INSERT INTO salaries VALUES(10001,88958,'1986-06-26','9999-01-01'); INSERT INTO salaries VALUES(10003,43311,'2001-12-01','9999-01-01'); INSERT INTO salaries VALUES(10004,70698,'1986-12-01','1995-12-01'); INSERT INTO salaries VALUES(10004,74057,'1995-12-01','9999-01-01'); INSERT INTO salaries VALUES(10006,43311,'2001-08-02','9999-01-01'); INSERT INTO salaries VALUES(10007,88070,'2002-02-07','9999-01-01'); INSERT INTO titles VALUES(10001,'Senior Engineer','1986-06-26','9999-01-01'); INSERT INTO titles VALUES(10003,'Senior Engineer','2001-12-01','9999-01-01'); INSERT INTO titles VALUES(10004,'Engineer','1986-12-01','1995-12-01'); INSERT INTO titles VALUES(10004,'Senior Engineer','1995-12-01','9999-01-01'); INSERT INTO titles VALUES(10006,'Senior Engineer','2001-08-02','9999-01-01'); INSERT INTO titles VALUES(10007,'Senior Staff','1996-02-11','9999-01-01');
（3）查询结果
select a.title,avg(b.salary) from titles a left join salaries b on a.emp_no=b.emp_no where a.to_date='9999-01-01' and b.to_date='9999-01-01' group by a.title;
17.获取当前（to_date='9999-01-01'）薪水第二多的员工的emp_no以及其对应的薪水salary
（1）创建表格
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`));
（2）查询结果
select emp_no,salary from salaries where to_date='9999-01-01' order by salary desc limit 1,1;
18.查找当前薪水(to_date='9999-01-01')排名第二多的员工编号emp_no、薪水salary、last_name以及first_name，你可以不使用order by完成吗
（1）创建表
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`));
（2）查询结果
select a.emp_no,a.salary,b.last_name,b.first_name from salaries a left join employees b on a.emp_no=b.emp_no where a.to_date='9999-01-01' and a.salary = (select max(salary) from salaries where salary != (select max(salary) from salaries));
19.查找所有员工的last_name和first_name以及对应的dept_name，也包括暂时没有分配部门的员工
（1）创建表
CREATE TABLE `departments` ( `dept_no` char(4) NOT NULL, `dept_name` varchar(40) NOT NULL, PRIMARY KEY (`dept_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 `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`));
（2）插入数据
INSERT INTO departments VALUES('d001','Marketing');
INSERT INTO departments VALUES('d002','Finance');
INSERT INTO departments VALUES('d003','Human Resources');
INSERT INTO departments VALUES('d004','Production');
INSERT INTO departments VALUES('d005','Development');
INSERT INTO departments VALUES('d006','Quality Management');
INSERT INTO dept_emp VALUES(10001,'d001','1986-06-26','9999-01-01');
INSERT INTO dept_emp VALUES(10002,'d001','1996-08-03','9999-01-01');
INSERT INTO dept_emp VALUES(10003,'d004','1995-12-03','9999-01-01');
INSERT INTO dept_emp VALUES(10004,'d004','1986-12-01','9999-01-01');
INSERT INTO dept_emp VALUES(10005,'d003','1989-09-12','9999-01-01');
INSERT INTO dept_emp VALUES(10006,'d002','1990-08-05','9999-01-01');
INSERT INTO dept_emp VALUES(10007,'d005','1989-02-10','9999-01-01');
INSERT INTO dept_emp VALUES(10008,'d005','1998-03-11','2000-07-31');
INSERT INTO dept_emp VALUES(10009,'d006','1985-02-18','9999-01-01');
INSERT INTO dept_emp VALUES(10010,'d005','1996-11-24','2000-06-26');
INSERT INTO dept_emp VALUES(10010,'d006','2000-06-26','9999-01-01');
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');
（3）查询结果
select a.last_name,a.first_name,c.dept_name from employees a left join dept_emp b on a.emp_no = b.emp_no left join departments c on b.dept_no = c.dept_no;
20.查找员工编号emp_no为10001其自入职以来的薪水salary涨幅(总共涨了多少)growth(可能有多次涨薪，没有降薪)
（1）创建表格
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`));
（2）查询结果
select MAX(salary)- MIN(salary) as growth from salaries where emp_no=10001;
21.查找所有员工自入职以来的薪水涨幅情况，给出员工编号emp_no以及其对应的薪水涨幅growth，并按照growth进行升序
（注:可能有employees表和salaries表里存在记录的员工，有对应的员工编号和涨薪记录，但是已经离职了，离职的员工salaries表的最新的to_date!='9999-01-01'，这样的数据不显示在查找结果里面）
（1）创建表格
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, -- '入职时间' PRIMARY KEY (`emp_no`)); CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL, `salary` int(11) NOT NULL, `from_date` date NOT NULL, -- '一条薪水记录开始时间' `to_date` date NOT NULL, -- '一条薪水记录结束时间' PRIMARY KEY (`emp_no`,`from_date`));
不会
22.统计各个部门的工资记录数，给出部门编码dept_no、部门名称dept_name以及部门在salaries表里面有多少条记录sum
（1）创建表格
CREATE TABLE `departments` ( `dept_no` char(4) NOT NULL, `dept_name` varchar(40) NOT NULL, PRIMARY KEY (`dept_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 `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`));
（2）查询结果
select a.dept_no,a.dept_name,count(*) as 'sum' from departments a left join dept_emp b on a.dept_no = b.dept_no left join salaries c on b.emp_no = c.emp_no group by a.dept_no
23.对所有员工的当前(to_date='9999-01-01')薪水按照salary进行按照1-N的排名，相同salary并列且按照emp_no升序排列
（1）创建表格
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`));
（2）查询结果
select emp_no,salary,dense_rank() over (order by salary desc) as rank from salaries where to_date='9999-01-01'
