我正在尝试在phpmyadmin中创建以下表格,我收到上述错误
-- --------------------------------------------------- -- Create County Table -- --------------------------------------------------- CREATE TABLE IF NOT EXISTS SecretGarden.County ( co_id varchar (8) NOT NULL UNIQUE , co_desc VARCHAR(50), st_date DATE, end_date DATE, PRIMARY KEY (co_id) ); -- --------------------------------------------------- -- Create Login Table -- --------------------------------------------------- CREATE TABLE IF NOT EXISTS SecretGarden.login ( email_id varchar (100) NOT NULL UNIQUE, password VARCHAR(50), st_date DATE, end_date DATE, PRIMARY KEY (email_id) ); -- --------------------------------------------------- -- Create Address Table -- --------------------------------------------------- CREATE TABLE IF NOT EXISTS SecretGarden.address ( add_id int (8) NOT NULL AUTO_INCREMENT , address1 VARCHAR(50), address2 VARCHAR(50), town VARCHAR(50), co_id varchar (8), st_date DATE, end_date DATE, PRIMARY KEY (add_id), CONSTRAINT fk_coid FOREIGN KEY (co_id) REFERENCES county(co_id)); -- --------------------------------------------------- -- Create StaffRole Table -- --------------------------------------------------- CREATE TABLE IF NOT EXISTS SecretGarden.staffrole ( staffrole_id int (8) NOT NULL AUTO_INCREMENT , staffrole_desc VARCHAR(50), st_date DATE, end_date DATE, PRIMARY KEY (staffrole_id));
以上所有内容都运行完美,但当我尝试创建下一个表时,我得到了上述错误
-- --------------------------------------------------- -- Create Staff Table -- --------------------------------------------------- CREATE TABLE IF NOT EXISTS SecretGarden.staff ( staff_id int (8) NOT NULL AUTO_INCREMENT , fname VARCHAR(50), lname VARCHAR(50), email_id varchar (100), home_no INT (20), mobile_no INT (20), add_id INT (8), staffrole_id INT (8), st_date DATE, end_date DATE, PRIMARY KEY (staff_id), CONSTRAINT fk_emailid FOREIGN KEY (email_id) REFERENCES SecretGarden.login(email_id), CONSTRAINT fk_addid FOREIGN KEY (add_id) REFERENCES SecretGarden.address(add_id), CONSTRAINT fk_staffrole FOREIGN KEY (staffrole_id) REFERENCES SecretGarden.staffrole(staffrole_id));
任何帮助将非常感激
雷切
答案 0 :(得分:0)
问题是我没有给每个约束一个单独的名字 我没有意识到它们对于整个数据库来说都是独一无二的