mysql建表是comment问题

Question

问题遇到的现象和发生背景

mysql8.0版本建表时，使用comment标注注释
发生错误无法使用

问题相关代码，请勿粘贴截图

CREATE DATABASE IF NOT EXISTS school;
-- 创建一个school数据库
USE school;

-- 创建年级表
DROP TABLE IF EXISTS grade;
CREATE TABLE grade(
gradeid INT(11) NOT NULL AUTO_INCREMENT COMMENT ‘年级编号’,
gradename VARCHAR(50) NOT NULL COMMENT ‘年级名称’,
PRIMARY KEY (gradeid)
) ENGINE=INNODB AUTO_INCREMENT = 6 DEFAULT CHARSET = utf8;

-- 插入年级数据
INSERT INTO grade (gradeid,gradename) VALUES(1,‘大一’),(2,‘大二’),(3,‘大三’),(4,‘大四’),(5,‘预科班’);

-- 创建成绩表
DROP TABLE IF EXISTS result;
CREATE TABLE result(
studentno INT(4) NOT NULL COMMENT ‘学号’,
subjectno INT(4) NOT NULL COMMENT ‘课程编号’,
examdate DATETIME NOT NULL COMMENT ‘考试日期’,
studentresult INT (4) NOT NULL COMMENT ‘考试成绩’,
KEY subjectno (subjectno)
)ENGINE = INNODB DEFAULT CHARSET = utf8;

-- 插入成绩数据 这里仅插入了一组，其余自行添加
INSERT INTO result(studentno,subjectno,examdate,studentresult)
VALUES
(1000,1,‘2013-11-11 16:00:00’,85),
(1000,2,‘2013-11-12 16:00:00’,70),
(1000,3,‘2013-11-11 09:00:00’,68),
(1000,4,‘2013-11-13 16:00:00’,98),
(1000,5,‘2013-11-14 16:00:00’,58);

-- 创建学生表
DROP TABLE IF EXISTS student;
CREATE TABLE student(
studentno INT(4) NOT NULL COMMENT ‘学号’,
loginpwd VARCHAR(20) DEFAULT NULL,
studentname VARCHAR(20) DEFAULT NULL COMMENT ‘学生姓名’,
sex TINYINT(1) DEFAULT NULL COMMENT ‘性别，0或1’,
gradeid INT(11) DEFAULT NULL COMMENT ‘年级编号’,
phone VARCHAR(50) NOT NULL COMMENT ‘联系电话，允许为空’,
address VARCHAR(255) NOT NULL COMMENT ‘地址，允许为空’,
borndate DATETIME DEFAULT NULL COMMENT ‘出生时间’,
email VARCHAR (50) NOT NULL COMMENT ‘邮箱账号允许为空’,
identitycard VARCHAR(18) DEFAULT NULL COMMENT ‘身份证号’,
PRIMARY KEY (studentno),
UNIQUE KEY identitycard(identitycard),
KEY email (email)
)ENGINE=MYISAM DEFAULT CHARSET=utf8;

-- 插入学生数据 其余自行添加 这里只添加了2行
INSERT INTO student (studentno,loginpwd,studentname,sex,gradeid,phone,address,borndate,email,identitycard)
VALUES
(1000,‘123456’,‘张伟’,0,2,‘13800001234’,‘北京朝阳’,‘1980-1-1’,‘[e-mail]’,‘123456198001011234’),
(1001,‘123456’,‘赵强’,1,3,‘13800002222’,‘广东深圳’,‘1990-1-1’,‘[e-mail]’,‘123456199001011233’);

-- 创建科目表
DROP TABLE IF EXISTS subject;
CREATE TABLE subject(
subjectnoINT(11) NOT NULL AUTO_INCREMENT COMMENT ‘课程编号’,
subjectname VARCHAR(50) DEFAULT NULL COMMENT ‘课程名称’,
classhour INT(4) DEFAULT NULL COMMENT ‘学时’,
gradeid INT(4) DEFAULT NULL COMMENT ‘年级编号’,
PRIMARY KEY (subjectno)
)ENGINE = INNODB AUTO_INCREMENT = 19 DEFAULT CHARSET = utf8;

-- 插入科目数据
INSERT INTO subject(subjectno,subjectname,classhour,gradeid)VALUES
(1,‘高等数学-1’,110,1),
(2,‘高等数学-2’,110,2),
(3,‘高等数学-3’,100,3),
(4,‘高等数学-4’,130,4),
(5,‘C语言-1’,110,1),
(6,‘C语言-2’,110,2),
(7,‘C语言-3’,100,3),
(8,‘C语言-4’,130,4),
(9,‘Java程序设计-1’,110,1),
(10,‘Java程序设计-2’,110,2),
(11,‘Java程序设计-3’,100,3),
(12,‘Java程序设计-4’,130,4),
(13,‘数据库结构-1’,110,1),
(14,‘数据库结构-2’,110,2),
(15,‘数据库结构-3’,100,3),
(16,‘数据库结构-4’,130,4),
(17,‘C#基础’,130,1);

运行结果及报错内容

Error Code: 1064. You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '‘年级编号’, gradename VARCHAR(50) NOT NULL COMMENT ‘年级名称’, P' at line 2

我的解答思路和尝试过的方法

我想要达到的结果

成功建表并插入数据

Answer 1

你所有的引号都是全角的中文标点，全部改成半角