SQL創建表 - 錯誤

Question

我目前使用phpmyadmin來創建許多不同的表，但這一個似乎是造成問題。這可能是這部分代碼的錯誤，或者可能來自於我的代碼中進一步引用Staff的內容？

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 - 無法創建表 'h_h.staff'（錯誤：150）（詳細資料...）

Answer 1

從錯誤，這是非常清楚，這個錯誤是由於外鍵約束。你可以嘗試禁用外鍵約束，然後創建你的表，並最終啓用它？

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

我想你應該創建這樣的外鍵索引

... 

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) 
);