牛客网sql实战
2020-02-25 本文已影响0人
笨鸡
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
