Question

我正在尝试创建表，然后显示以下错误。但我找不到错误。我是Mysql的新手请帮助我..

SQL查询：

--
-- Dumping data for table `files`
--
-- --------------------------------------------------------
--
-- Table structure for table `post`
--
CREATE TABLE IF NOT EXISTS  `offer` (    
 offer_id INT( 9 ) NOT NULL AUTO_INCREMENT ,
 offer_category VARCHAR( 50 ) CHARACTER SET utf8 NOT NULL ,
 offer_type VARCHAR( 50 ) CHARACTER SET utf8 NOT NULL ,
 sub_category_id INT( 9 ) NOT NULL ,
 category_id INT( 9 ) NOT NULL ,
 state_id INT( 9 ) NOT NULL ,
 city_id INT( 9 ) NOT NULL ,
 location_id INT( 9 ) NOT NULL ,
 file_id INT( 9 ) NOT NULL ,
 title VARCHAR( 50 ) CHARACTER,
 description TEXT CHARACTER,
 date TIMESTAMP NOT NULL DEFAULT CURRENT_TIMESTAMP ,
 exp_date TIMESTAMP NOT NULL DEFAULT CURRENT_TIMESTAMP ,
 shop_id INT( 9 ) NOT NULL ,
 status TINYINT( 1 ) NOT NULL ,
PRIMARY KEY (  offer_id )
) ENGINE = INNODB DEFAULT CHARSET = latin1 AUTO_INCREMENT =63;

错误是，

MySQL said: Documentation

        #1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '  
          `description` text CHARACTER,
            `date` timestamp NOT NULL DEFAULT CURRENT_TI' at line 22

Answer 1

问题：

title VARCHAR（50）字符：VARCHAR（50）就足够了，MySQL不支持这种语法。
description文字字符：使用一个TEXT或CHAR
两列不能有CURRENT_TIMESTAMP

试试这个

CREATE TABLE IF NOT EXISTS  `offer` (

 `offer_id` INT( 9 ) NOT NULL AUTO_INCREMENT ,
 `offer_category` VARCHAR( 50 ) CHARACTER SET utf8 NOT NULL ,
 `offer_type` VARCHAR( 50 ) CHARACTER SET utf8 NOT NULL ,
 `sub_category_id` INT( 9 ) NOT NULL ,
 `category_id` INT( 9 ) NOT NULL ,
 `state_id` INT( 9 ) NOT NULL ,
 `city_id` INT( 9 ) NOT NULL ,
 `location_id` INT( 9 ) NOT NULL ,
 `file_id` INT( 9 ) NOT NULL ,
 `title` VARCHAR( 50 ) ,
 `description` TEXT ,
 `date` TIMESTAMP NOT NULL DEFAULT CURRENT_TIMESTAMP ,
 `exp_date` TIMESTAMP NOT NULL DEFAULT '0000-00-00 00:00:00' ,
 `shop_id` INT( 9 ) NOT NULL ,
 `status` TINYINT( 1 ) NOT NULL ,
PRIMARY KEY (  `offer_id` )
) ENGINE = INNODB DEFAULT CHARSET = latin1 AUTO_INCREMENT =63;

MySQL＃1064 - 您的SQL语法出错

1 个答案: