牛客网sql实战参考答案（mysql版）：1-15

1、查找最晚入职员工的所有信息，为了减轻入门难度，目前所有的数据里员工入职的日期都不是同一天(sqlite里面的注释为--,mysql为comment)

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`));

select *

from employees

order by hire_date desc

limit 0, 1;

2、查找入职员工时间排名倒数第三的员工所有信息，为了减轻入门难度，目前所有的数据里员工入职的日期都不是同一天

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

order by hire_date desc

limit 1 offset 2;

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)

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`));

select s.emp_no, s.salary, s.from_date, s.to_date, d.dept_no

from salaries s

inner join dept_manager d on s.emp_no = d.emp_no

where d.to_date='9999-01-01' and s.to_date='9999-01-01'

order by s.emp_no;

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)

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`));

select e.last_name, e.first_name, d.dept_no

from employees e

inner join dept_emp d on d.emp_no = e.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)

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`));

select e.last_name, e.first_name, d.dept_no

from employees e

left join dept_emp d on d.emp_no = e.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)

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, s.salary

from employees e

inner join salaries s on s.emp_no = e.emp_no

where e.hire_date = s.from_date

order by e.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

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, count(emp_no) t

from salaries

group by emp_no

having t > 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

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 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，输出结果按照dept_no升序排列(请注意，同一个人可能有多条薪水情况记录)

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`));

select m.dept_no, m.emp_no, s.salary

from dept_manager m

inner join salaries s on s.emp_no = m.emp_no

where m.to_date='9999-01-01' and s.to_date='9999-01-01'

order by m.dept_no;

10、获取所有非manager的员工emp_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 `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`));

如插入为:

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');

select e.emp_no

from employees e

left join dept_manager m on e.emp_no = m.emp_no

where m.dept_no is null;

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。

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, m.emp_no manager_no

from dept_emp e, dept_manager m

where e.emp_no != m.emp_no

and e.dept_no = m.dept_no

and m.to_date = '9999-01-01';

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，按照部门编号升序排列。

【这题做错过】

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');

参考别人的答案

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;

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);

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(title) t

from titles

group by title

having t >= 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。

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 count(*) >= 2;

只需在count中加入distinct '指定列'就可以在计算行数时去掉指定列中重复的值：在count函数计数时去掉重复的emp_no

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

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 *

from employees

where emp_no % 2 = 1

and last_name != "Mary"

order by hire_date desc;

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