牛客网sql实战

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 ORDER BY hire_date desc limit 1;  

2.查找入职员工时间排名倒数第三的员工所有信息

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 
    emp_no = (
        select a.emp_no from employees a join employees b 
            on a.hire_date < b.hire_date 
            GROUP BY a.emp_no 
            HAVING count(a.emp_no) = 2);

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(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`));

这题有坑
select b.*, a.dept_no from dept_manager a join salaries b on a.emp_no = b.emp_no
   and b.to_date = '9999-01-01' and a.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 e.last_name, e.first_name, d.dept_no from dept_emp d join employees e 
    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，也包括展示没有分配具体部门的员工

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`));

写右连不通过。。。
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

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 join salaries s
    on e.emp_no = s.emp_no and e.hire_date = s.from_date
    order by e.emp_no desc;

7.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 count(emp_no) > 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

select DISTINCT salary from salaries 
    where to_date = '9999-01-01' order by salary desc;

(性能更优)
select salary from salaries
    where to_date = '9999-01-01' 
    group by salary order by salary desc;

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 
    dept_manager d join salaries s
    on d.emp_no = s.emp_no 
    and s.to_date = '9999-01-01' and d.to_date = '9999-01-01';

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`));

select emp_no from (
    select e.emp_no, d.dept_no as dept from employees e left join dept_manager d
    on d.emp_no = e.emp_no
) where dept is null;

SELECT e.emp_no FROM employees e LEFT JOIN dept_manager d
    ON e.emp_no = d.emp_no
    WHERE dept_no IS NULL

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 de join dept_manager dm
    on de.dept_no = dm.dept_no 
    and de.emp_no <> dm.emp_no
    and de.to_date = '9999-01-01'
    and dm.to_date = '9999-01-01'

12.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, d.emp_no, s.salary
    from dept_emp d left 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 dept_no HAVING MAX(salary);

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 emp_no&1 and last_name <> 'Mary'
    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 t.title, avg(salary) as avg
    from titles t join salaries s
    on t.emp_no = s.emp_no
    and s.to_date = '9999-01-01'
    and t.to_date = '9999-01-01'
    group by 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'
    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 e.emp_no, s.salary, e.last_name, e.first_name 
    from employees e right join    
        (select b.emp_no, b.salary
        from salaries a join salaries b
        on a.to_date = '9999-01-01' 
                and b.to_date = '9999-01-01'
                and a.salary > b.salary
        GROUP BY b.emp_no 
        HAVING count(b.emp_no) = 1) as s
    on e.emp_no = s.emp_no;

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 de.emp_no = e.emp_no left join departments d
    on d.dept_no = de.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;

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 e.emp_no, (a.salary - b.salary) as growth 
    from employees e join salaries a
    on e.emp_no = a.emp_no and a.to_date = '9999-01-01'
    join salaries b
    on e.emp_no = b.emp_no and e.hire_date = b.from_date
    order by growth;

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, d.dept_name, count(d.dept_no)
    from dept_emp de join departments d
    on de.dept_no = d.dept_no
    join salaries s
    on de.emp_no = s.emp_no
    group by d.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

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 s1 join salaries s2
    on s1.salary <= s2.salary 
    and s1.to_date = '9999-01-01' 
    and s2.to_date = '9999-01-01'
    group by s1.emp_no order by rank;

24.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 de.dept_no, de.emp_no, s.salary
    from employees e join dept_emp de
    on de.emp_no = e.emp_no
    join salaries s
    on e.emp_no = s.emp_no and s.to_date = '9999-01-01'
    where e.emp_no not in(
        select d.emp_no from dept_manager d
            WHERE d.to_date = '9999-01-01'
    )

25.获取员工其当前的薪水比其manager当前薪水还高的相关信息，当前表示to_date='9999-01-01',结果第一列给出员工的emp_no，第二列给出其manager的manager_no，第三列给出该员工当前的薪水emp_salary,第四列给该员工对应的manager当前的薪水manager_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 `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 es.emp_no as emp_no, ds.emp_no as manager_no, es.salary as emp_salary, ds.salary as manager_salary from 
    (select s.emp_no, de.dept_no, s.salary from dept_emp de join salaries s
        on de.emp_no = s.emp_no and s.to_date = '9999-01-01') as es join
    (select s.emp_no, dm.dept_no, s.salary from dept_manager dm join salaries s
        on dm.emp_no = s.emp_no and s.to_date = '9999-01-01') as ds
    on es.dept_no = ds.dept_no and es.salary > ds.salary

26.汇总各个部门当前员工的title类型的分配数目，结果给出部门编号dept_no、dept_name、其当前员工所有的title以及该类型title对应的数目count

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 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 d.dept_no, d.dept_name, t.title, count(t.title) as count
    from departments d join dept_emp de
    on d.dept_no = de.dept_no 
    and de.to_date = '9999-01-01'
    join titles t 
    on de.emp_no = t.emp_no
    and t.to_date = '9999-01-01'
    group by de.dept_no, t.title

27.给出每个员工每年薪水涨幅超过5000的员工编号emp_no、薪水变更开始日期from_date以及薪水涨幅值salary_growth，并按照salary_growth逆序排列。提示：在sqlite中获取datetime时间对应的年份函数为strftime('%Y', to_date)

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.from_date, (s1.salary - s2.salary) as salary_growth from
    salaries s1 join salaries s2
    on s1.emp_no = s2.emp_no 
    and strftime('%Y', s1.to_date) - strftime('%Y', s2.to_date) = 1
    and s1.salary - s2.salary > 5000
    order by salary_growth desc

28.查找描述信息中包括robot的电影对应的分类名称以及电影数目，而且还需要该分类对应电影数量>=5部

CREATE TABLE IF NOT EXISTS film (
film_id smallint(5)  NOT NULL DEFAULT '0',
title varchar(255) NOT NULL,
description text,
PRIMARY KEY (film_id));
CREATE TABLE category  (
category_id  tinyint(3)  NOT NULL ,
name  varchar(25) NOT NULL, `last_update` timestamp,
PRIMARY KEY ( category_id ));
CREATE TABLE film_category  (
film_id  smallint(5)  NOT NULL,
category_id  tinyint(3)  NOT NULL, `last_update` timestamp);

select c.name, count(f.film_id) as count
    from category c join film_category fc
    on c.category_id = fc.category_id
    join film f
    on fc.film_id = f.film_id
    and f.description like '%robot%'
    group by c.name
    having count >= 2;

29.使用join查询方式找出没有分类的电影id以及名称

select f.film_id, f.title from
    film f left join film_category fc
    on f.film_id = fc.film_id
    where fc.category_id is null

30.使用子查询的方式找出属于Action分类的所有电影对应的title,description

select title, description from film
    where film_id in (
        select film_id from film_category
            where category_id = (
                select category_id from category
                    where name = 'Action'
            )
    )

31.获取select * from employees对应的执行计划

explain select * from employees

32.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

select last_name||" "||first_name as name  from employees

MySQL
select concat_ws(" ", last_name, first_name) as Name from employees

33.创建一个actor表，包含如下列信息

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')))

34.对于表actor批量插入如下数据

insert into actor values 
    (1,'PENELOPE','GUINESS','2006-02-15 12:34:33'),
    (2,'NICK','WAHLBERG','2006-02-15 12:34:33');

35.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

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')))

INSERT or IGNORE INTO actor VALUES ('3', 'ED', 'CHASE', '2006-02-15 12:34:33')

MySQL
INSERT IGNORE INTO actor VALUES ('3', 'ED', 'CHASE', '2006-02-15 12:34:33')

36.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：

sqlite
create table actor_name as select first_name,last_name from actor;

mysql
select first_name, last_name into actor_name from actor

42.删除emp_no重复的记录，只保留最小的id对应的记录。

CREATE TABLE IF NOT EXISTS titles_test (
id int(11) not null primary key,
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);

delete from titles_test where id not in
    (
        select min(id) from titles_test group by emp_no
    )

43.将所有to_date为9999-01-01的全部更新为NULL,且 from_date更新为2001-01-01。

CREATE TABLE IF NOT EXISTS titles_test (
id int(11) not null primary key,
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);

update titles_test set to_date = NULL, from_date = '2001-01-01' 
    where to_date = '9999-01-01'

44.将id=5以及emp_no=10001的行数据替换成id=5以及emp_no=10005,其他数据保持不变，使用replace实现。

CREATE TABLE IF NOT EXISTS titles_test (
id int(11) not null primary key,
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);

replace into titles_test values(5, 10005, 'Senior Engineer', '1986-06-26', '9999-01-01');

45.将titles_test表名修改为titles_2017。

CREATE TABLE IF NOT EXISTS titles_test (
id int(11) not null primary key,
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);

RENAME TABLE titles_test to titles_2017;
 
ALTER TABLE titles_test RENAME TO titles_2017;

46.在audit表上创建外键约束，其emp_no对应employees_test表的主键id。

CREATE TABLE employees_test(
ID INT PRIMARY KEY NOT NULL,
NAME TEXT NOT NULL,
AGE INT NOT NULL,
ADDRESS CHAR(50),
SALARY REAL
);

CREATE TABLE audit(
EMP_no INT NOT NULL,
create_date datetime NOT NULL
);

DROP TABLE audit;
CREATE TABLE audit(
    EMP_no INT NOT NULL,
    create_date datetime NOT NULL,
    FOREIGN KEY(EMP_no) REFERENCES employees_test(ID));

47.存在如下的视图：

create view emp_v as select * from employees where emp_no >10005;
如何获取emp_v和employees有相同的数据？

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 emp_v;

48.将所有获取奖金的员工当前的薪水增加10%。

create table emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);
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`));

update salaries set salary = salary * (1.1) 
    where to_date = '9999-01-01'
    and emp_no in (select emp_no from emp_bonus);

49.针对库中的所有表生成select count(*)对应的SQL语句

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 emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);
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 `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 count(*) from " || name || ";" AS cnts
    FROM sqlite_master WHERE type = 'table'

mysql
select table_name from information_schema.tables where table_schema='shop' ;

50.将employees表中的所有员工的last_name和first_name通过(')连接起来。

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 as name  from employees

MySQL
select concat_ws("'", last_name, first_name) as Name from employees

51.查找字符串'10,A,B' 中逗号','出现的次数cnt。

select length('10,A,B') -length(replace('10,A,B',",","")) as cnt

52.获取Employees中的first_name，查询按照first_name最后两个字母，按照升序进行排列

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 first_name as f FROM employees
    order by substr(f, length(f) - 1)

53.按照dept_no进行汇总，属于同一个部门的emp_no按照逗号进行连接，结果给出dept_no以及连接出的结果employees

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`));

select dept_no, group_concat(emp_no) as employees
    from dept_emp group by dept_no;

select dept_no, group_concat(emp_no SEPARATOR ',') 
    from dept_emp group by dept_no;

54.查找排除当前最大、最小salary之后的员工的平均工资avg_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 avg(salary) as avg_salary
    from salaries
        where salary not in (select max(salary) from salaries where to_date = '9999-01-01')
        and (select min(salary) from salaries where to_date = '9999-01-01')
        and to_date = '9999-01-01';

55.分页查询employees表，每5行一页，返回第2页的数据

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 limit 5, 5;

56.获取所有员工的emp_no、部门编号dept_no以及对应的bonus类型btype和received ，没有分配具体的员工不显示

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`));
create table emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);

select e.emp_no, de.dept_no, eb.btype, eb.recevied
    from employees e inner join dept_emp de 
    on e.emp_no = de.emp_no
    left join emp_bonus eb
    on e.emp_no = eb.emp_no

57.使用含有关键字exists查找未分配具体部门的员工的所有信息。

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`));
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`));

select * from employees where not exists 
    (    
        select emp_no from dept_emp where emp_no = employees.emp_no
    )

58.存在如下的视图：

create view emp_v as select * from employees where emp_no >10005;
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`));
获取employees中的行数据，且这些行也存在于emp_v中。注意不能使用intersect关键字。

select * from emp_v;

59.获取有奖金的员工相关信息。

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`));
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 emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);
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`));
给出emp_no、first_name、last_name、奖金类型btype、对应的当前薪水情况
salary以及奖金金额bonus。 bonus类型btype为1其奖金为薪水salary的10%，
btype为2其奖金为薪水的20%，其他类型均为薪水的30%。 当前薪水表示to_date='9999-01-01'

select e.emp_no, e.first_name, e.last_name, eb.btype, s.salary,
    (
        CASE eb.btype WHEN 1 THEN s.salary * 0.1
                      WHEN 2 THEN s.salary * 0.2
            ELSE s.salary * 0.3
        END
    ) AS bonus
    from employees e join emp_bonus eb
    on e.emp_no = eb.emp_no
    join salaries s
    on e.emp_no = s.emp_no and s.to_date = '9999-01-01'

60.按照salary的累计和running_total，其中running_total为前两个员工的salary累计和，其他以此类推。 具体结果如下Demo展示。。

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 s1.emp_no, s1.salary, sum(s2.salary) AS running_total
    from salaries s1 join salaries s2
    on s1.emp_no >= s2.emp_no 
    and s1.to_date = '9999-01-01'
    and s2.to_date = '9999-01-01'
    group by s1.emp_no
    order by s1.emp_no;

61.对于employees表中，给出奇数行的first_name

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 e1.first_name from employees e1 
    where (
        select count(*) from employees e2 where e1.first_name >= e2.first_name
    )&1