MySQL笔试题练习及答案(二)

第十一题
题目描述
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。
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));

如插入：
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’);

select
e.emp_no as emp_no,
m.emp_no as manager_no
from dept_emp e
join dept_manager m
on e.dept_no = m.dept_no
where e.emp_no <> m.emp_no
and e.to_date = '9999-01-01'
and m.to_date = '9999-01-01';

第十二题
题目描述
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，按照部门编号升序排列。
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));
如插入：
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’);

select uni.dept_no, uni.emp_no, max_salary.salary
from
    (select d.dept_no, s.emp_no, s.salary
     from dept_emp d join salaries s
     on d.emp_no = s.emp_no
     and d.to_date = '9999-01-01'
     and s.to_date = '9999-01-01'  
    ) as uni, /* 部门编号,员工编号,当前薪水 */
    (select d.dept_no, max(s.salary) as salary
     from dept_emp d join salaries s
     on d.emp_no = s.emp_no
     and d.to_date = '9999-01-01'
     and s.to_date = '9999-01-01'  
     group by d.dept_no
    ) as max_salary /* 部门编号,当前最高薪水 */
where uni.salary = max_salary.salary
and uni.dept_no = max_salary.dept_no
order by uni.dept_no;

第十三题
题目描述
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);
如插入：
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’);

select 
title,
count(emp_no) t
from titles 
group by title 
having t >= 2;

第十四题
题目描述
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(即emp_no重复的title不计算，title对应的数目t不增加)。
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);
如插入：
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’);

select 
title,
count(distinct emp_no) t
from titles
group by title 
hAVing t >= 2;

第十五题
题目描述
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

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));

如插入：
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’);

select 
emp_no,
birth_date,
first_name,
last_name,
gender,
hire_date
from employees 
where emp_no % 2 = 1 and last_name != 'Mary'
order by hire_date desc;
-- 本题注意的要点:1. order by 在 where 后面.2.多个条件用and连接,3. 字符串需要加''

第十六题
题目描述
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。
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);

如插入：
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’);

select 
t.title,
avg(s.salary) as avg
from salaries s join titles t 
on s.emp_no = t.emp_no 
and t.to_date='9999-01-01'
and s.to_date='9999-01-01'
group by t.title;

第十七题
题目描述
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
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));

select 
emp_no,
salary
from salaries
where to_date='9999-01-01'
order by salary desc
limit 1,1;

--  严谨版本
select 
emp_no, 
salary 
from salaries
where to_date = '9999-01-01' 
and salary = (select distinct salary from salaries order by salary desc limit 1,1)
-- 避免了2个问题：
-- (1) 首先这样可以解决多个人工资相同的问题；
-- (2) 另外，筛选出第二多的工资时要注意distinct salary，否则不能选出第二多的工资。

第十八题
题目描述
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
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));

select 
e.emp_no,
max(s.salary) as salary,
e.last_name,
e.first_name
from employees e
join salaries s
on e.emp_no  = s.emp_no 
where s.to_date = '9999-01-01'
and s.salary < (select max(salary) from salaries where to_date = '9999-01-01');

第十九题
题目描述
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
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));

如插入：
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’);

select 
e.last_name,
e.first_name,
t.dept_name
from employees e
left join 
(select emp_no,dept_name from departments dep join dept_emp d on dep.dept_no = d.dept_no) t
on e.emp_no = t.emp_no;

第二十题
题目描述
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)
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));

select (
(select salary from salaries where emp_no = 10001 order by to_date desc limit 1) -
(select salary from salaries where emp_no = 10001 order by to_date asc limit 1)
) growth;

在这里插入代码片

更多相关链接
MySQL笔试题练习及答案(一)
MySQL笔试题练习及答案(二)
MySQL笔试题练习及答案(三)