MySQL数据库题库练习

-- 构造一个触发器audit_log，在向employees_test表中插入一条数据的时候，触发插入相关的数据到audit中。
drop table if exists audit;
drop table if exists employees_test;
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,
NAME TEXT NOT NULL
);
select * from employees_test;
select * from audit;

-- 正确答案
-- 创建触发器
create TRIGGER audit_log
after -- after/before/
insert -- insert/update/delete
on employees_test -- on table_name
for each row -- for each row
begin
insert into audit values(new.id,new.NAME); # 触发事件 使用新列数据
end
SHOW TRIGGERS; -- 查询触发器
-- 向 employees_test 表插入数据操作，然后查询 audit表是否有数据:
INSERT INTO employees_test (ID,NAME,AGE,ADDRESS,SALARY)VALUES (1,'Paul',32,'California',20000.00 );
INSERT INTO employees_test (ID,NAME,AGE,ADDRESS,SALARY)VALUES (4,'gd',33,'California',20000.00 );
INSERT INTO employees_test (ID,NAME,AGE,ADDRESS,SALARY)VALUES (6,'Pagdgul',34,'California',20000.00 );
INSERT INTO employees_test (ID,NAME,AGE,ADDRESS,SALARY)VALUES (5,'dgd',35,'California',20000.00 );
select * from audit;
DROP TRIGGER IF EXISTS audit_log ; -- 删除触发器

-- 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
drop table if exists `dept_emp` ;
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`));
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');
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
输出格式:
dept_no employees
d001 10001,10002
d002 10006
d003 10005
d004 10003,10004
d005 10007,10008,10010
d006 10009,10010
select * from dept_emp;

-- 正确答案
select dept_no,GROUP_CONCAT(emp_no) as employees from dept_emp GROUP BY dept_no

-- 请你查找在职员工自入职以来的薪水涨幅情况，给出在职员工编号emp_no以及其对应的薪水涨幅growth，并按照growth进行升序，以上例子输出为（注: to_date为薪资调整某个结束日期，或者为离职日期，to_date='9999-01-01'时，表示依然在职，无后续调整记录）
-- 系数 A
drop table if exists `employees` ;
drop table if exists `salaries` ;
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`));
INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','2001-06-22');
INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1999-08-03');
INSERT INTO salaries VALUES(10001,85097,'2001-06-22','2002-06-22');
INSERT INTO salaries VALUES(10001,88958,'2002-06-22','9999-01-01');
INSERT INTO salaries VALUES(10002,72527,'1999-08-03','2000-08-02');
INSERT INTO salaries VALUES(10002,72527,'2000-08-02','2001-08-02');
SELECT * from employees;
SELECT * from salaries;

-- 正确答案
SELECT
t1.emp_no,
(t2.salary - t1.salary) AS growth
FROM
(
SELECT
e.emp_no,
s.salary
FROM
employees e
LEFT JOIN salaries s ON e.emp_no = s.emp_no
AND e.hire_date = s.from_date
) t1
INNER JOIN (
SELECT
em.emp_no,
sa.salary
FROM
employees em
LEFT JOIN salaries sa ON em.emp_no = sa.emp_no
WHERE
sa.to_date = '9999-01-01'
) t2
WHERE
t1.emp_no = t2.emp_no ORDER BY growth

-- id为用户主键id，number代表积分情况，让你写一个sql查询积分表里面出现三次以及三次以上的积分，查询结果如下:
注意：若有多个符合条件的number，则按number升序排序输出。
drop table if exists grade;
CREATE TABLE `grade` (
`id` int(4) NOT NULL,
`number` int(4) NOT NULL,
PRIMARY KEY (`id`));
INSERT INTO grade VALUES
(1,111),
(2,333),
(3,111),
(4,111),
(5,333);
select * from grade;

-- 正确答案 默认就是升序排列
select number from grade GROUP BY number HAVING COUNT(number)>=3

-- 查找排除在职(to_date = '9999-01-01' )员工的最大、最小salary之后，其他的在职员工的平均工资avg_salary。
drop table if exists `salaries` ;
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` float(11,3) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
INSERT INTO salaries VALUES(10001,85097,'2001-06-22','2002-06-22');
INSERT INTO salaries VALUES(10001,88958,'2002-06-22','9999-01-01');
INSERT INTO salaries VALUES(10002,72527,'2001-08-02','9999-01-01');
INSERT INTO salaries VALUES(10003,43699,'2000-12-01','2001-12-01');
INSERT INTO salaries VALUES(10003,43311,'2001-12-01','9999-01-01');
INSERT INTO salaries VALUES(10004,70698,'2000-11-27','2001-11-27');
INSERT INTO salaries VALUES(10004,74057,'2001-11-27','9999-01-01');
select * from salaries;

-- 正确答案
select (SUM(salary)-MAX(salary)-MIN(salary))/(COUNT(1)-2) AS 平均工资 from salaries where to_date in('9999-01-01');

-- 请将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。(注：sqllite,字符串拼接为 || 符号，不支持concat函数，mysql支持concat函数)。
drop table if exists `employees` ;
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`));
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 ;
-- 正确答案
select CONCAT(last_name,' ',first_name) from employees ;

-- 汇总各个部门当前员工的title类型的分配数目，即结果给出部门编号dept_no、dept_name、其部门下所有的员工的title以及该类型title对应的数目count，结果按照dept_no升序排序，dept_no一样的再按title升序排序
-- 系数 A
drop table if exists `departments` ;
drop table if exists `dept_emp` ;
drop table if exists titles ;
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 titles (
`emp_no` int(11) NOT NULL,
`title` varchar(50) NOT NULL,
`from_date` date NOT NULL,
`to_date` date DEFAULT NULL);
INSERT INTO departments VALUES('d001','Marketing');
INSERT INTO departments VALUES('d002','Finance');
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,'d002','1995-12-03','9999-01-01');
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');
SELECT * from departments ;
SELECT * from dept_emp ;
SELECT * from titles ;

-- 正确答案
-- 1
SELECT
pts.dept_no,
pts.dept_name,
les.title,
(
SELECT
count(title)
FROM
titles t
WHERE
t.emp_no = dep.emp_no
) AS count1
FROM
dept_emp dep
LEFT JOIN departments pts ON dep.dept_no = pts.dept_no
LEFT JOIN titles les ON dep.emp_no = les.emp_no
where les.to_date='9999-01-01'
and dep.to_date='9999-01-01'
ORDER BY
pts.dept_no,
count1
-- 2
select d.dept_no,
d.dept_name,
t.title,
count(t.title)as count
from departments d,dept_emp de,titles t
where de.emp_no=t.emp_no
and de.dept_no=d.dept_no
and de.to_date='9999-01-01'
and t.to_date='9999-01-01'
group by d.dept_no,t.title