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MySQL Workbench:查询错误(1064):第1行“ VISIBLE”附近的语法错误

为什么VISIBLE下面会引起问题的任何想法?

CREATE TABLE IF NOT EXISTS `setting` (
  `uuid` INT(10) NOT NULL,
  `type` VARCHAR(255) NOT NULL,
  `code` VARCHAR(255) NOT NULL COMMENT 'An unique name.',
  `value` MEDIUMTEXT NULL DEFAULT NULL,
  `comment` LONGTEXT NULL DEFAULT NULL,
  `created_on` INT UNSIGNED NOT NULL,
  `updated_on` INT UNSIGNED NOT NULL,
  PRIMARY KEY (`uuid`))
ENGINE = MyISAM
DEFAULT CHARACTER SET = utf8;

CREATE UNIQUE INDEX `name_UNIQUE` ON `setting` (`code` ASC) VISIBLE;

CREATE UNIQUE INDEX `uuid_UNIQUE` ON `setting` (`uuid` ASC) VISIBLE;
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错误:

CREATE UNIQUE INDEX name_UNIQUE …

mysql mysql-workbench

10
推荐指数
1
解决办法
4775
查看次数

MySQL错误1064语法,但一切似乎都很好

我正在使用xampp控制面板,然后从那里开始apache和mysql的过程。然后我转到mysql工作台,服务器状态似乎正常,这是一些信息

Host: Windows-PC
Socket: C:/xampp/mysql/mysql.sock
Port: 3306
Version 10.1.31-MariaDB mariadb.org binary distribution
Compiled For: Win32(32)
Configuratin File: unknown
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然后每次当我尝试为我的虚拟架构添加外键时:

 ALTER TABLE `puppies`.`animals` 
 ADD INDEX `Breed_idx` (`BreedID` ASC) VISIBLE;
 ;
 ALTER TABLE `puppies`.`animals` 
 ADD CONSTRAINT `Breed`
 FOREIGN KEY (`BreedID`)
 REFERENCES `puppies`.`breeds` (`Breed`)
 ON DELETE NO ACTION
 ON UPDATE NO ACTION;
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我收到以下错误

 ERROR 1064: You have an error in your SQL syntax; check the manual that 
 corresponds to your MariaDB server version for the right syntax to use near 
 '' at line …
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mysql xampp mariadb

7
推荐指数
3
解决办法
9438
查看次数

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