这是错误
ical subscribe access key
附近使用正确的语法, created_on
datetime DEFAULT NULL,
在第20行“更新”
ERREUR 请求SQL:
-- --------------------------------------------------------
--
-- Structure de la table `pec_calendars`
--
CREATE TABLE `pec_calendars` (
`id` INT( 11 ) UNSIGNED NOT NULL ,
`type` ENUM( 'user', 'group', 'url' ) DEFAULT 'user',
`user_id` INT( 11 ) UNSIGNED DEFAULT NULL ,
`name` VARCHAR( 100 ) DEFAULT NULL ,
`description` TEXT,
`color` VARCHAR( 7 ) DEFAULT NULL ,
`admin_id` INT( 11 ) DEFAULT NULL ,
`status` ENUM( 'on', 'off' ) DEFAULT 'on',
`show_in_list` ENUM( '0', '1' ) DEFAULT NULL ,
`public` TINYINT( 3 ) UNSIGNED DEFAULT '0',
`reminder_message_email` TEXT,
`reminder_message_popup` TEXT,
`access_key` VARCHAR( 32 ) DEFAULT NULL COMMENT AS `ical subscribe access key` ,
`created_on` DATETIME DEFAULT NULL ,
`updated_on` DATETIME DEFAULT NULL
) ENGINE = INNODB DEFAULT CHARSET = utf8;
答案 0 :(得分:1)
猜猜你的查询应该是:
CREATE TABLE pec_calendars
(
id INT(11) UNSIGNED NOT NULL,
type ENUM('user', 'group', 'url') DEFAULT 'user',
user_id INT(11) UNSIGNED DEFAULT NULL,
name VARCHAR(100) DEFAULT NULL,
description TEXT,
color VARCHAR(7) DEFAULT NULL,
admin_id INT(11) DEFAULT NULL,
status ENUM('on', 'off') DEFAULT 'on',
show_in_list ENUM('0', '1') DEFAULT NULL,
public TINYINT(3) UNSIGNED DEFAULT '0',
reminder_message_email TEXT,
reminder_message_popup TEXT,
access_key VARCHAR(32) DEFAULT NULL,
created_on DATETIME DEFAULT NULL,
updated_on DATETIME DEFAULT NULL
)
engine = innodb
DEFAULT charset = utf8;