Mysql 我正试图用sql实现这段代码，但还是遇到了一个错误，我真的不知道为什么

您的代码中存在多个问题：

• 很少有语法是不正确的
• mysql中应该使用

  int（6）

  而不是

  number（6）

• varchar

  必须具有数据大小。-<代码>varchar（200）
• 必须在创建引用表后创建外键

示例模式中正确语法的几个示例如下：

create table student (
student_ID number(6) primary key,
student_name varchar (25) unique ,
address varchar ,
sex char, 
manager_ID number(6) REFERENCES maneger (manager_ID) , 
birth_date date NOT NULL 
);
create table Classes (
class_ID number(6) primary key ,
student_ID number (6)REFERENCES student (student_ID), FOREIGN KEY
);
create table manager (
manager_ID number (6) primary key,
manager_name char ,
address varchar,
Teacher_ID number (6) FOREIGN KEY, REFERENCES teacher (Teacher_ID),
sex char
);
create table teacher (
Teacher_ID number (6), primary key,
teacher_name varchar, 
address varchar , 
class_ID number (6)  FOREIGN KEY REFERENCES Classes (class_ID) ,
subject_name varchar REFERENCES subject (subject_name)
);
create table subject (
subject_name varchar  primary key,
marks number 
);
create learn (
student_ID number (6) REFERENCES student (student_ID), 
subject_name varchar REFERENCES subject (subject_name) , 
constraint PK_student primary key (student_ID,subject_name)
);
create table teach (
student_ID number (6)REFERENCES student (student_ID) , 
Teacher_ID number (6) REFERENCES teacher (Teacher_ID),
constraint PK_teach primary key (student_ID,Teacher_ID)
);

---

具有正确的表定义。请使用

alter table\u name add constraint添加外键…

如果您告诉我们错误，则会有所帮助。您需要先创建引用表，然后才能在student中正确引用表和拼写管理器。

create table student 
(
    student_ID int(6) primary key,
    student_name varchar (25) unique ,
    address varchar(200) ,
    sex char(10), 
    manager_ID int(6),
    birth_date date NOT NULL 
);

create table Classes 
(
    class_ID int(6) primary key ,
    student_ID int (6) references student (student_ID) 
);