牛客网SQL编程
1，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`));
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
select * from employees order by hire_date desc limit 2,1
SQL中limit的用法
select * from tableName limit i,n
# i：为查询结果的索引值(默认从0开始)，当i=0时可省略i
# n：为查询结果返回的数量
# i与n之间使用英文逗号","隔开
3， 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`));
select salaries.*, dept_manager.dept_no from salaries
inner join dept_manager on dept_manager.emp_no = salaries.emp_no
where/and dept_manager.to_date='9999-01-01' and salaries.to_date='9999-01-01';
4， 查找所有已经分配部门的员工的last_name和first_name以及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`));
select employees.last_name, employees.first_name, dept_emp.dept_no
from employees inner join dept_emp on dept_emp.emp_no = employees.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
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
INNER JOIN 两边表同时有对应的数据，即任何一边缺失数据就不显示。
LEFT JOIN 会读取左边数据表的全部数据，即便右边表无对应数据。
RIGHT JOIN 会读取右边数据表的全部数据，即便左边表无对应数据。
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
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`));
方法一：利用 INNER JOIN 连接两张表
SELECT e.emp_no, s.salary FROM employees AS e INNER JOIN salaries AS s
ON e.emp_no = s.emp_no AND e.hire_date = s.from_date
ORDER BY e.emp_no DESC
方法二：直接用逗号并列查询两张表
SELECT e.emp_no, s.salary FROM employees AS e, salaries AS s
WHERE e.emp_no = s.emp_no AND e.hire_date = s.from_date
ORDER BY e.emp_no DESC
- 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`));
SELECT emp_no, COUNT(emp_no) AS 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;
select salary from salaries where to_date='9999-01-01' group by salary ORDER BY salary DESC;
distinct是将所有查询的字段进行对比去重，所有字段都完全相同才会去重；必须放在查询字段开头进行查询
group by 根据字段进行去重，字段相同就会去重
对于distinct与group by的使用:
1、当对系统的性能高并数据量大时使用group by
2、当对系统的性能不高时使用数据量少时两者皆可
3、尽量使用group by
9，获取所有部门当前manager的当前薪水情况，给出dept_no, emp_no以及salary，当前表示to_date=‘9999-01-01’
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 salaries AS s INNER JOIN dept_manager AS d
ON d.emp_no = s.emp_no
AND d.to_date = '9999-01-01'
AND s.to_date = '9999-01-01'
- 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`));
方法一：使用NOT IN选出在employees但不在dept_manager中的emp_no记录
SELECT emp_no FROM employees
WHERE emp_no NOT IN (SELECT emp_no FROM dept_manager)
方法二：先使用LEFT JOIN连接两张表，再从此表中选出dept_no值为NULL对应的emp_no记录
SELECT employees.emp_no FROM employees LEFT JOIN dept_manager
ON employees.emp_no = dept_manager.emp_no
WHERE dept_no IS NULL
方法三：EXCEPT 集合差运算； UNION 集合并运算； INTERSECT 集合交运算
SELECT employees.emp_no FROM salaries
EXCEPT
SELECT dept_manager.emp_no FROM dept_manager;
11， 获取所有员工当前的manager，如果当前的manager是自己的话结果不显示，当前表示to_date=‘9999-01-01’。
结果第一列给出当前员工的emp_no,第二列给出其manager对应的manager_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 `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`));
SELECT de.emp_no, dm.emp_no AS manager_no
FROM dept_emp AS de INNER JOIN dept_manager AS dm
ON de.dept_no = dm.dept_no
WHERE dm.to_date = '9999-01-01' AND de.to_date = '9999-01-01' AND de.emp_no <> dm.emp_no
e.emp_no <> m.emp_no ===== e.emp_no!=m.emp_no
- 获取所有部门中当前员工薪水最高的相关信息，给出dept_no, emp_no以及其对应的salary
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`));
SELECT d.dept_no, s.emp_no, MAX(s.salary) AS salary
FROM salaries AS s INNER JOIN dept_emp As d
ON d.emp_no = s.emp_no
WHERE d.to_date = '9999-01-01' AND s.to_date = '9999-01-01'
GROUP BY d.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);
SELECT title, COUNT(*) AS 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。
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);
SELECT title, COUNT(DISTINCT emp_no) AS t FROM titles
GROUP BY title HAVING t >= 2
15,
查找employees表所有emp_no为奇数，且last_name不为Mary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`));
select * from employees
where last_name != 'Mary'
and emp_no % 2 = 1
order by hire_date desc
16, 统计出当前各个title类型对应的员工当前（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);
select title,avg(salary) as avg
from titles a
inner join salaries b
on a.emp_no=b.emp_no and 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
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'
GROUP BY salary -------------- distinct salary，否则不能选出第二多的工资。
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，不准使用order by
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 s.emp_no , max(s.salary) , e.last_name, e.first_name
from salaries s, employees e
where s.emp_no = e.emp_no and s.to_date='9999-01-01'
and s.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
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`));
select e.last_name, e.first_name, d.dept_name
from employees e left join dept_emp de
on e.emp_no = de.emp_no
left join departments d
on de.dept_no = d.dept_no
20， 查找员工编号emp_no为10001其自入职以来的薪水salary涨幅值growth
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 max(salary)-min(salary) as growth from salaries where emp_no = '10001'
严谨写法：
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)
) AS growth
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
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 a.emp_no, (b.salary - c.salary) as growth
from
employees as a
inner join salaries as b
on a.emp_no = b.emp_no and b.to_date = '9999-01-01'
inner join salaries as c
on a.emp_no = c.emp_no and a.hire_date = c.from_date
order by growth asc
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以及次数sum
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`));
select d.dept_no ,dept.dept_name,count(salary)
from salaries s ,dept_emp d,departments dept
where s.emp_no = d.emp_no and d.dept_no=dept.dept_no
group by dept.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
牛客答案讨论1111111
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 s1.emp_no, s1.salary, COUNT(DISTINCT s2.salary) AS rank
FROM salaries AS s1, salaries AS s2
WHERE s1.to_date = '9999-01-01'
AND s2.to_date = '9999-01-01'
AND s1.salary <= s2.salary
GROUP BY s1.emp_no
ORDER BY s1.salary DESC, s1.emp_no ASC
- 获取所有非manager员工当前的薪水情况，给出dept_no、emp_no以及salary ，当前表示to_date=‘9999-01-01’
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 `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`));
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, s.emp_no, s.salary
from dept_emp d, salaries s
where s.emp_no NOT IN (SELECT emp_no FROM dept_manager dm where dm.to_date='9999-01-01' )
and d.emp_no = s.emp_no and s.to_date='9999-01-01'
. 将employees表的所有员工的last_name和first_name拼接起来作为Name，中间以一个空格区分
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`));
MySQL、SQL Server、Oracle等数据库支持CONCAT方法，
而本题所用的SQLite数据库只支持用连接符号"||"来连接字符串
CONCAT方法：
select CONCAT(CONCAT(last_name," "),first_name) as name from employees
或者
select CONCAT(last_name," "，first_name) as name from employees
本题中使用：
select last_name||" "||first_name as name from employees
. 创建一个actor表，包含如下列信息
列表 类型 是否为NULL 含义
actor_id smallint(5) not null 主键id
first_name varchar(45) not null 名字
last_name varchar(45) not null 姓氏
last_update timestamp not null 最后更新时间，默认是系统的当前时间
CREATE TABLE actor
(
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime('now','localtime')) --,
-- PRIMARY KEY(actor_id)
)
.
对于表actor批量插入如下数据
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime('now','localtime')))
actor_id first_name last_name last_update
1 PENELOPE GUINESS 2006-02-15 12:34:33
2 NICK WAHLBERG 2006-02-15 12:34:33
insert into 表名 values (数据1), (数据2)
注意： 引号
insert into actor
values (1, 'PENELOPE', 'GUINESS', '2006-02-15 12:34:33'),
(2, 'NICK', 'WAHLBERG', '2006-02-15 12:34:33')
. 对于表actor批量插入如下数据,如果数据已经存在，请忽略，不使用replace操作
insert or ignore into
insert or ignore into actor
values (3, 'ED', 'CHASE', '2006-02-15 12:34:33')
. 从旧表里导入数据到新表中
对于如下表actor，其对应的数据为:
actor_id first_name last_name last_update
1 PENELOPE GUINESS 2006-02-15 12:34:33
2 NICK WAHLBERG 2006-02-15 12:34:33
创建一个actor_name表，将actor表中的所有first_name以及last_name导入改表。 actor_name表结构如下：
列表 类型 是否为NULL 含义
first_name varchar(45) not null 名字
last_name varchar(45) not null 姓氏
题目使用的是sqlite3，可以这么做：
create table actor_name as
select first_name,last_name from actor;
如果是mysql，那么as可以去掉，也可以不去掉，例如：
create table actor_name
select first_name,last_name from actor;
. 创建唯一索引和普通索引
针对如下表actor结构创建索引：
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime('now','localtime')))
对first_name创建唯一索引uniq_idx_firstname，对last_name创建普通索引idx_lastname
create unique index uniq_idx_firstname on actor(first_name);
create index idx_lastname on actor(last_name);
. 创建视图
针对actor表创建视图actor_name_view，只包含first_name以及last_name两列，并对这两列重新命名，first_name为first_name_v，last_name修改为last_name_v：
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime('now','localtime')))
create view actor_name_view (first_name_v, last_name_v) as
select first_name,last_name from actor
.
针对salaries表emp_no字段创建索引idx_emp_no，查询emp_no为10005, 使用强制索引。
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 index idx_emp_no on salaries(emp_no);