Question

您好我正在尝试创建一些表，当一个新用户在我的php网站中注册并且我正在尝试执行事务中的最后一个查询时它会给我以下错误：#1005 - Can't create table 'user_39.records' (errno: 150) 有人可以指出我做错了什么吗？我不太熟悉外键，这是我第一次使用它们。谢谢。

数据库表：

private function createDriversTable($database) {
    $query = "CREATE TABLE IF NOT EXISTS $database.drivers (
      `id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
      `name` varchar(255) NOT NULL,
      `surname` varchar(255) NOT NULL,
      `car_type` varchar(255) NOT NULL,
      `circulation_num` varchar(255) NOT NULL,
      `special_card_num` varchar(255) NOT NULL,
      `special_card_num_exp` date NOT NULL,
      `drivers_license_num` varchar(255) NOT NULL,
      `drivers_license_exp` date NOT NULL,
      `id_num` varchar(255) NOT NULL,
      `vat` varchar(255) NOT NULL,
      `address` varchar(255) NOT NULL,
      `mobile` bigint(20) NOT NULL,
      `radiotaxi_code` bigint(20) NOT NULL,
      `languages` varchar(255) NOT NULL,
      `email` varchar(255) NOT NULL,
      PRIMARY KEY (`id`)
    ) ENGINE=InnoDB DEFAULT CHARSET=utf8 COMMENT='Driver records' AUTO_INCREMENT=1";

    return $query;
}

private function createClientsTable($database) {
    $query = "CREATE TABLE IF NOT EXISTS $database.clients (
      `id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
      `name` varchar(255) NOT NULL,
      `address` varchar(255) NOT NULL,
      `region` varchar(255) NOT NULL,
      `vat` varchar(255) NOT NULL COMMENT 'AFM',
      `tax_office` varchar(255) NOT NULL COMMENT 'DOY',
      `phone` bigint(20) NOT NULL,
      `mobile` bigint(20) NOT NULL,
      `email` varchar(255) NOT NULL,
      `notes` text NOT NULL,
      PRIMARY KEY (`id`)
    ) ENGINE=InnoDB DEFAULT CHARSET=utf8 COMMENT='Client records' AUTO_INCREMENT=1"; 

    return $query;
}

private function createCarsTable($database) {
    $query = "CREATE TABLE IF NOT EXISTS $database.cars (
      `id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
      `oil_change` date NOT NULL,
      `oil_filter` date NOT NULL,
      `petrol_filter` date NOT NULL,
      `air_filter` date NOT NULL,
      `cabin_filter` date NOT NULL,
      `carbon_filter` date NOT NULL,
      `front_breaks` date NOT NULL,
      `rear_breaks` date NOT NULL,
      `front_disc_breaks` date NOT NULL,
      `rear_disc_breaks` date NOT NULL,
      `break_fluids` date NOT NULL,
      `gear_oil` date NOT NULL,
      `gear_controller` date NOT NULL,
      `gear_filter` date NOT NULL,
      `tires` date NOT NULL,
      `kteo` date NOT NULL,
      `freon` date NOT NULL,
      `freon_filter` date NOT NULL,
      `steering_fluids` date NOT NULL,
      `axle_oil` date NOT NULL,
      `notes` text NOT NULL,
      PRIMARY KEY (`id`)
    ) ENGINE=InnoDB DEFAULT CHARSET=utf8 COMMENT='Car records' AUTO_INCREMENT=1";

    return $query;
}

private function createRecordsTable($database) {
    $query = "CREATE TABLE IF NOT EXISTS $database.records (
      `id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
      `date` date NOT NULL,
      `time` time NOT NULL,
      `driver_id` bigint(20) NOT NULL,
      `client_id` bigint(20) NOT NULL,
      `car_id` bigint(20) NOT NULL,
      `room_num` int(11) NOT NULL,
      `departure` text NOT NULL,
      `course` text NOT NULL,
      `destination` text NOT NULL,
      `arrives_from` text NOT NULL,
      `arrival_info` text NOT NULL,
      `route_type` bigint(20) NOT NULL,
      `payment_method` int(11) NOT NULL,
      `total_cost` float NOT NULL,
      `expenses` float NOT NULL,
      `profit` float NOT NULL,
      `notes` text NOT NULL,
      PRIMARY KEY (`id`),
      FOREIGN KEY (`driver_id`) REFERENCES $database.drivers(`id`),
      FOREIGN KEY (`client_id`) REFERENCES $database.clients(`id`),
      FOREIGN KEY (`car_id`) REFERENCES $database.cars(`id`)
    ) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=1";

    return $query;      
}

Answer 1

您缺少driver_id和client_id

的结果表中的无符号类型

Answer 2

要成为另一个表中的 FOREIGN KEY ，您必须在另一个表上创建索引。并且为了能够可靠地引用特定行，因此它应该是UNIQUE索引。否则，如果您有重复值，则引用表将无法识别它引用的行。即使InnoDB允许您在非唯一索引上创建关系，也可能不是您正在寻找的行为

Answer 3

首先，您的表格未正常化。尝试快速搜索“数据库规范化”：它将对您有很大帮助。

另外，就目前而言，我们无法回答您的问题，因为我们缺少一些关键信息：相关的表格结构。

目前，请看一下这些要点：

是否为BIGINT类型的user_39.drivers（id）？
是否为BIGINT类型的user_39.clients（id）？
是否为BIGINT类型的user_39.cars（id）？

我猜其中一个不是BIGINT，数据类型差异阻止你创建外键

SQL使用外键创建表查询

3 个答案: