Question

当我在phpMyAdmin和Sequel上运行查询时，我收到以下错误

＃1064 - 您的SQL语法出错;查看与您的MySQL服务器版本对应的手册，以便在'id Int UNSIGNED NOT NULL AUTO_INCREMENT附近使用正确的语法，登录Char（128）NOT NULL，passw'在第3行。

我无法弄清楚此查询的错误。 Mysql在5.5.25版本上运行。

CREATE TABLE user
(
  id Int UNSIGNED NOT NULL AUTO_INCREMENT,
  login Char(128) NOT NULL,
  password Char(128) NOT NULL,
  emailaddress Char(128),
  alternateemailaddress Char(128),
  officephone Char(20),
  officefax Char(20),
  mobilephone Char(20),
  client Int UNSIGNED,
  facility Int UNSIGNED,
  user_status Int UNSIGNED NOT NULL DEFAULT 0,
  valid_from Datetime NOT NULL,
  valid_to Datetime NOT NULL,
  last_login Datetime,
  last_login_from Char(48),
  modified_by Int NOT NULL,
  modified_time Datetime,

  PRIMARY KEY (id),
  FOREIGN KEY client REFERENCES client (id) ON DELETE CASCADE,
  FOREIGN KEY facility REFERENCES facility (id) ON DELETE CASCADE
 
 ) ENGINE = InnoDB
  AUTO_INCREMENT = 0
;

Answer 1

在编写表和字段的名称时应该使用后台，以避免MySQL将其解析为保留关键字。您还应该使用包装外键名称的括号。工作查询应如下所示：

CREATE TABLE `user`
(
  `id` Int UNSIGNED NOT NULL AUTO_INCREMENT,
  `login` Char(128) NOT NULL,
  `password` Char(128) NOT NULL,
  `emailaddress` Char(128),
  `alternateemailaddress` Char(128),
  `officephone` Char(20),
  `officefax` Char(20),
  `mobilephone` Char(20),
  `client` Int UNSIGNED,
  `facility` Int UNSIGNED,
  `user_status` Int UNSIGNED NOT NULL DEFAULT 0,
  `valid_from` Datetime NOT NULL,
  `valid_to` Datetime NOT NULL,
  `last_login` Datetime,
  `last_login_from` Char(48),
  `modified_by` Int NOT NULL,
  `modified_time` Datetime,

  PRIMARY KEY (`id`),
  FOREIGN KEY (`client`) REFERENCES `client` (`id`) ON DELETE CASCADE,
  FOREIGN KEY (`facility`) REFERENCES `facility` (`id`) ON DELETE CASCADE

 ) ENGINE = InnoDB
  AUTO_INCREMENT = 0
;

Answer 2

这对我有用。
查看我的成功SQLFiddle

我怀疑你的输入中有一些奇怪的空白字符，也许是标签。

创建查询时出现SQL语法错误

2 个答案: