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SQL牛客网SQL实战(1-10)题解

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SQL第1-10题解答(SQL)

本文章尽量用简洁的代码去完成题目要求,话不多说,上代码,后期会不断更新的。

1.查找最晚入职员工的所有信息

题目描述: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
where hire_date =(select max(hire_date) from employees)

2.查找入职员工时间排名倒数第三的员工所有信息

题目描述: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.查找各个部门当前领导当前薪水详情以及其对应部门编号dept_no

题目描述: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:请以salaries表为主表进行查询,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.*,d.dept_no from salaries s,dept_manager d
where d.to_date = '9999-01-01'
and s.to_date='9999-01-01'
and s.emp_no = d.emp_no;

4.查找所有已经分配部门的员工的last_name和first_name

题目描述: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 last_name,first_name,dept_no from employees e,dept_emp d
where e.emp_no = d.emp_no;

5.查找所有员工的last_name和first_name

题目描述: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 last_name,first_name,dept_no from employees e
left join dept_emp d
on e.emp_no = d.emp_no;

6.查找所有员工入职时候的薪水情况

题目描述: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 emp_no,salary from salaries
group by emp_no having salary<=min(salary)
order by emp_no desc;

7.查找薪水涨幅超过15次变动的员工

题目描述: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(*) t from salaries
group by emp_no
having count(*)>15;

8.找出所有员工当前具体的薪水salary情况

题目描述: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 s
where s.to_date='9999-01-01'
order by salary desc

9.获取所有部门当前manager的当前薪水情况

题目描述: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)
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 d.dept_no, d.emp_no, s.salary from dept_manager d,salaries s
where d.emp_no = s.emp_no
and d.to_date='9999-01-01'
and s.to_date='9999-01-01'

10.获取所有非manager的员工emp_no

题目描述: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 emp_no from employees
where emp_no not in (
select e.emp_no
from employees e,dept_manager d
where e.emp_no = d.emp_no
)

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