我的数据库中有两个实体都需要地址。客户可以有多个地址。餐厅每个ID可以有一个地址。我应该如何组织关系?我目前有一个地址表,它引用了它所属的用户,但它也可以属于一家餐厅。
地址:
CREATE TABLE IF NOT EXISTS grabatakeaway.address (
`address_id` int(8) UNSIGNED AUTO_INCREMENT NOT NULL PRIMARY KEY,
`address` text NOT NULL,
`city` varchar(128) NOT NULL,
`state_province` varchar(128),
`zip_post` varchar(32) NOT NULL,
`username` varchar(32) NOT NULL,
FOREIGN KEY (username) REFERENCES grabatakeaway.user(username)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
用户:
CREATE TABLE IF NOT EXISTS grabatakeaway.user (
`username` varchar(32) NOT NULL PRIMARY KEY,
`password` varchar(128) NOT NULL,
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
餐厅:
CREATE TABLE IF NOT EXISTS grabatakeaway.restaurant (
`restaurant_id` int(8) UNSIGNED AUTO_INCREMENT NOT NULL PRIMARY KEY,
`address_id` int(8) UNSIGNED NOT NULL,
FOREIGN KEY (address_id) REFERENCES grabatakeaway.address(address_id)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
您需要一个'through'表。
CREATE TABLE IF NOT EXISTS (user_address
address_id int,
user_name varchar(32),
FOREIGN KEY (username) REFERENCES grabatakeaway.user(username),
FOREIGN KEY (address_id) REFERENCES grabatakeaway.address(address_id),
PRIMARY KEY (user_id,user_name))
地址表中不需要用户名。
CREATE TABLE IF NOT EXISTS grabatakeaway.address (
`address_id` int(8) UNSIGNED AUTO_INCREMENT NOT NULL PRIMARY KEY,
`address` text NOT NULL,
`city` varchar(128) NOT NULL,
`state_province` varchar(128),
`zip_post` varchar(32) NOT NULL,
FOREIGN KEY (username) REFERENCES grabatakeaway.user(username)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
我还可以建议您为users表设置一个自动递增的id字段吗?这将导致它的索引变小,直通表的索引也变小。