SQL创建表 - 错误

Question

我目前使用phpmyadmin来创建许多不同的表，但这一个似乎是造成问题。这可能是这部分代码的错误，或者可能来自于我的代码中进一步引用Staff的内容？

SQL query: 

CREATE TABLE Staff(

Staff_ID INTEGER(5) PRIMARY KEY AUTO_INCREMENT , 
Local_ID INTEGER(5) , 
First_name VARCHAR(20) NOT NULL , 
Last_name VARCHAR(20) NOT NULL , 
Address_line_1 VARCHAR(30) NOT NULL , 
Address_line_2 VARCHAR(30) NOT NULL , 
City VARCHAR(20) NOT NULL , 
Post_Code VARCHAR(8) NOT NULL , 
Email VARCHAR(30) NOT NULL , 
Telephone INTEGER NOT NULL , 
Date_employed DATE, 
Salary DECIMAL, 
Sales_ID INTEGER(5) , 
Manager_ID INTEGER(5) , 
Development_ID INTEGER(5) , 
FOREIGN KEY (Local_ID) REFERENCES LocalOffice(Local_ID) , 
FOREIGN KEY (Sales_ID) REFERENCES Sales_Advisors(Sales_ID) , 
FOREIGN KEY (Manager_ID) REFERENCES Site_Manager(Manager_ID) , 
FOREIGN KEY (Development_ID) REFERENCES Development(Development_ID) 
); 

MySQL表示：

1005 - 无法创建表 'h_h.staff'（错误：150）（详细资料...）

Answer 1

从错误，这是非常清楚，这个错误是由于外键约束。你可以尝试禁用外键约束，然后创建你的表，并最终启用它？

SET foreign_key_checks = 0; 

CREATE TABLE Staff(

Staff_ID INTEGER(5) PRIMARY KEY AUTO_INCREMENT , 
Local_ID INTEGER(5) , 
First_name VARCHAR(20) NOT NULL , 
Last_name VARCHAR(20) NOT NULL , 
Address_line_1 VARCHAR(30) NOT NULL , 
Address_line_2 VARCHAR(30) NOT NULL , 
City VARCHAR(20) NOT NULL , 
Post_Code VARCHAR(8) NOT NULL , 
Email VARCHAR(30) NOT NULL , 
Telephone INTEGER NOT NULL , 
Date_employed DATE, 
Salary DECIMAL, 
Sales_ID INTEGER(5) , 
Manager_ID INTEGER(5) , 
Development_ID INTEGER(5) , 
FOREIGN KEY (Local_ID) REFERENCES LocalOffice(Local_ID) , 
FOREIGN KEY (Sales_ID) REFERENCES Sales_Advisors(Sales_ID) , 
FOREIGN KEY (Manager_ID) REFERENCES Site_Manager(Manager_ID) , 
FOREIGN KEY (Development_ID) REFERENCES Development(Development_ID) 
); 

SET foreign_key_checks = 1; 

Answer 2

我想你应该创建这样的外键索引

... 

Development_ID INTEGER(5) , 
INDEX Local_ID, 
INDEX Sales_ID, 
INDEX Manager_ID, 
INDEX DEvelopment_ID, 
FOREIGN KEY (Local_ID) REFERENCES LocalOffice(Local_ID) , 
FOREIGN KEY (Sales_ID) REFERENCES Sales_Advisors(Sales_ID) , 
FOREIGN KEY (Manager_ID) REFERENCES Site_Manager(Manager_ID) , 
FOREIGN KEY (Development_ID) REFERENCES Development(Development_ID) 
);