Question

我已将项目从XAMP移至VM-＆gt; Vagrant-＆gt; ubuntu / trusty64-＆gt;已安装（php5-mysql mysql-server mysql-client）

现在，当我尝试创建数据库和表时，我收到以下错误。

错误无法创建表properties，外键约束不正确。引用的表中没有索引，其中引用的列显示为第一列。

这是我的第一个脚本，所以我假设有一些错误但是当我使用XAMP时它都运行良好。我一直在阅读你使用的引擎和字符集，但我不太了解它或如何在流浪汉中编辑它。感谢所有帮助代码如下。

的MySQL

CREATE TABLE `usertypes`(
`id` INT NOT NULL AUTO_INCREMENT,
`type` CHAR(15),
`permissions` CHAR(15),
PRIMARY KEY (`id`)
);

INSERT INTO `usertypes`(`type`,`permissions`) VALUES ("Administrator", '{"Admin":1}' ), ("Staff", '{"Staff":1}'), ("Basic", '{"Basic":1}'), ("Pro", '{"Pro":1}'), ("Business", '{"Business":1}');

CREATE TABLE `users` (
`id` INT PRIMARY KEY AUTO_INCREMENT,
`username` VARCHAR (25) NOT NULL UNIQUE,
`password` VARCHAR (60) NOT NULL,
`usertype` INT NOT NULL DEFAULT 3,
`email` varchar(40) NOT NULL,
`authentication` varchar(32) NOT NULL,
`active` smallint(1) NOT NULL DEFAULT 0,
`newsletter` INT NOT NULL,
`banned` SMALLINT NOT NULL DEFAULT 0,
`user_since` DATE NOT NULL,
`listings` INT DEFAULT 0,
CONSTRAINT fk_usertype FOREIGN KEY (`usertype`) REFERENCES usertypes(`id`),
UNIQUE KEY `users` (`username`,`email`)
);

CREATE TABLE `personal` (
`id` INT PRIMARY KEY,
`first_name` VARCHAR (15) NULL,
`last_name` VARCHAR (15) NULL,
`mobile_phone` CHAR(10) NULL,
`city` CHAR(25) NULL,
`address` VARCHAR (100) NULL,
`postal_code` VARCHAR(6) NULL,
CONSTRAINT fk_personal FOREIGN KEY (`id`)
REFERENCES users(`id`)
);

/* Create Properties table */
CREATE TABLE `propertytypes`(
`id` INT NOT NULL AUTO_INCREMENT,
`type` CHAR(10),
PRIMARY KEY (`id`)
);

INSERT INTO `propertytypes`(`type`) VALUES ("house"), ("duplex"), ("apartment"), ("townhouse"), ("4 plex"),("6 plex"), ("room"), ("commercial"), ("gathering halls");

CREATE TABLE `utilities`(
`id` INT NOT NULL AUTO_INCREMENT,
`options` CHAR(10),
PRIMARY KEY (`id`)
);

INSERT INTO `utilities`(`options`) VALUES ("None"),("Heat"), ("Electricity"), ("Water"), ("TV"), ("Internet");

CREATE TABLE `features`(
`id` INT NOT NULL AUTO_INCREMENT,
`options` CHAR(10),
PRIMARY KEY (`id`)
);

INSERT INTO `features`(`options`) VALUES ("Coin Laundry"), ("Own Laundry"), ("Jetted Tub / Jacuzzi"), ("Gym"), ("Pool"), ("Security"), ("Balcony"), ("Elevator"), ("Hardwood Floors"), ("Fenced Backyard"), ("Dishwasher"), ("Air Conditioning"), ("Laminate Floors"), ("Fire Pit"), ("Fireplace"), ("Wheelchair Access"), ("Storage Lockers");

CREATE TABLE `properties` (
 `id` INT NOT NULL AUTO_INCREMENT,
 `owner` INT NOT NULL,
 `propertytype` INT NOT NULL,
 `address` varchar(255) NOT NULL,
 `postalcode` varchar(8) NOT NULL,
 `beds` INT(10),
 `baths` INT(10),
 `sqf` DECIMAL (6,2),
 `price` DECIMAL (6,2),
 `images` varchar(32),
 `listed` smallint(1) DEFAULT 2,
 `pets` smallint(1) DEFAULT 2,
 `kids` smallint(1) DEFAULT 2,
 `adults` smallint(1) DEFAULT 2,
 `utilities` INT NOT NULL DEFAULT 0,
 `features` INT NOT NULL DEFAULT 0,
 `parking` smallint(1) DEFAULT 2,
 `smokers` smallint(1) DEFAULT 2,
 `deposit` smallint(1) DEFAULT 0,
 `description` TEXT NOT NULL,
PRIMARY KEY (`id`),
CONSTRAINT fk_users FOREIGN KEY (`owner`)
REFERENCES users(`id`),
CONSTRAINT fk_propertytypes FOREIGN KEY (`propertytype`)
REFERENCES propertytypes(`id`),
CONSTRAINT fk_utilities FOREIGN KEY (`utilities`)
REFERENCES utilities(`options`),
CONSTRAINT fk_features FOREIGN KEY (`features`)
REFERENCES features(`options`)
);

Answer 1

这一直贯穿

查找说改变

的行

未正确指定列。另外，在插入过程中有2列有截断。

create schema ff; -- create a new db
use ff; -- now use that db

CREATE TABLE `usertypes`(
`id` INT NOT NULL AUTO_INCREMENT,
`type` CHAR(15),
`permissions` CHAR(15),
PRIMARY KEY (`id`)
);

INSERT INTO `usertypes`(`type`,`permissions`) VALUES ("Administrator", '{"Admin":1}' ), ("Staff", '{"Staff":1}'), ("Basic", '{"Basic":1}'), ("Pro", '{"Pro":1}'), ("Business", '{"Business":1}');

CREATE TABLE `users` (
`id` INT PRIMARY KEY AUTO_INCREMENT,
`username` VARCHAR (25) NOT NULL UNIQUE,
`password` VARCHAR (60) NOT NULL,
`usertype` INT NOT NULL DEFAULT 3,
`email` varchar(40) NOT NULL,
`authentication` varchar(32) NOT NULL,
`active` smallint(1) NOT NULL DEFAULT 0,
`newsletter` INT NOT NULL,
`banned` SMALLINT NOT NULL DEFAULT 0,
`user_since` DATE NOT NULL,
`listings` INT DEFAULT 0,
CONSTRAINT fk_usertype FOREIGN KEY (`usertype`) REFERENCES usertypes(`id`),
UNIQUE KEY `users` (`username`,`email`)
);

CREATE TABLE `personal` (
`id` INT PRIMARY KEY,
`first_name` VARCHAR (15) NULL,
`last_name` VARCHAR (15) NULL,
`mobile_phone` CHAR(10) NULL,
`city` CHAR(25) NULL,
`address` VARCHAR (100) NULL,
`postal_code` VARCHAR(6) NULL,
CONSTRAINT fk_personal FOREIGN KEY (`id`)
REFERENCES users(`id`)
);

/* Create Properties table */
-- drop table propertytypes;
CREATE TABLE `propertytypes`(
`id` INT NOT NULL AUTO_INCREMENT,
`type` CHAR(40), -- CHANGE MADE ************************
PRIMARY KEY (`id`)
);

INSERT INTO `propertytypes`(`type`) VALUES ("house"), ("duplex"), ("apartment"), ("townhouse"), ("4 plex"),("6 plex"), ("room"), ("commercial"), ("gathering halls");

-- drop table utilities;
CREATE TABLE `utilities`(
`id` INT NOT NULL AUTO_INCREMENT,
`options` CHAR(40), -- CHANGE MADE ************************
PRIMARY KEY (`id`)
);

INSERT INTO `utilities`(`options`) VALUES ("None"),("Heat"), ("Electricity"), ("Water"), ("TV"), ("Internet");

CREATE TABLE `features`(
`id` INT NOT NULL AUTO_INCREMENT,
`options` CHAR(40), -- CHANGE MADE ************************
PRIMARY KEY (`id`)
);

INSERT INTO `features`(`options`) VALUES ("Coin Laundry"), ("Own Laundry"), ("Jetted Tub / Jacuzzi"), ("Gym"), ("Pool"), ("Security"), ("Balcony"), ("Elevator"), ("Hardwood Floors"), ("Fenced Backyard"), ("Dishwasher"), ("Air Conditioning"), ("Laminate Floors"), ("Fire Pit"), ("Fireplace"), ("Wheelchair Access"), ("Storage Lockers");

CREATE TABLE `properties` (
 `id` INT NOT NULL AUTO_INCREMENT,
 `owner` INT NOT NULL,
 `propertytype` INT NOT NULL,
 `address` varchar(255) NOT NULL,
 `postalcode` varchar(8) NOT NULL,
 `beds` INT(10),
 `baths` INT(10),
 `sqf` DECIMAL (6,2),
 `price` DECIMAL (6,2),
 `images` varchar(32),
 `listed` smallint(1) DEFAULT 2,
 `pets` smallint(1) DEFAULT 2,
 `kids` smallint(1) DEFAULT 2,
 `adults` smallint(1) DEFAULT 2,
 `utilities` INT NOT NULL DEFAULT 0,
 `features` INT NOT NULL DEFAULT 0,
 `parking` smallint(1) DEFAULT 2,
 `smokers` smallint(1) DEFAULT 2,
 `deposit` smallint(1) DEFAULT 0,
 `description` TEXT NOT NULL,
PRIMARY KEY (`id`),

CONSTRAINT fk_users FOREIGN KEY (`owner`)
REFERENCES users(`id`),

CONSTRAINT fk_propertytypes FOREIGN KEY (`propertytype`)
REFERENCES propertytypes(`id`),

CONSTRAINT fk_utilities FOREIGN KEY (`utilities`)
REFERENCES utilities(`id`), -- CHANGE MADE ************************

CONSTRAINT fk_features FOREIGN KEY (`features`)
REFERENCES features(`id`) -- CHANGE MADE ************************
);

drop schema ff; -- drop test db (cleanup)

另外，如果DEFAULT 0与从1开始向上的FK约束发生冲突，请考虑删除它。坦率地说，我没有看到它的重点。

来自名为Using FOREIGN KEY Constraints

的mysql手册页

外键和引用键中的对应列必须有类似的数据类型。整数类型的大小和符号必须是相同。字符串类型的长度不必相同。对于非二进制（字符）字符串列，字符集和排序规则必须是一样的。

和

MySQL要求外键和引用键上的索引外键检查可以很快，不需要表扫描。在里面引用表时，必须有一个索引所在的外键列以相同顺序列为第一列。这样的如果不是，则会自动在引用表上创建索引存在。如果您创建，可能会稍后以静默方式删除此索引另一个可用于强制执行外键约束的索引。如果给定，则使用index_name，如前所述。

由于外键约束，在ubuntu / trusty64 php5-mysql上无法创建表

1 个答案: