无法在joomla 2.5中声明外键

Question

CREATE TABLE IF NOT EXISTS `#__web_projects` 
(
`id` int(11) NOT NULL AUTO_INCREMENT,
`type_website` varchar(30) COLLATE utf8_polish_ci NOT NULL,
`web_color` varchar(255) COLLATE utf8_polish_ci NOT NULL,
`web_fonts` varchar(255) COLLATE utf8_polish_ci NOT NULL,
`web_layout` text COLLATE utf8_polish_ci NOT NULL,
`web_menu` text COLLATE utf8_polish_ci NOT NULL,
`similar_web_sites` text COLLATE utf8_polish_ci NOT NULL,
`additional_info` text COLLATE utf8_polish_ci,
`about_company` text COLLATE utf8_polish_ci,
`offer` text COLLATE utf8_polish_ci,
`logo` varchar(255) COLLATE utf8_polish_ci NOT NULL,
`user_id` int(11) NOT NULL,
PRIMARY KEY(id),
FOREIGN KEY (`user_id`) REFERENCES `#__users`(id) ON DELETE CASCADE) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_polish_ci

这是我在joomla中的表声明，我总是得到这个表无法创建的错误。什么可能导致??

Answer 1

我想提一下，你应该让你使用user_id UNSIGNED！

我在Joomla中测试了你的SQL！ 2.5并且它完美无瑕（我无论如何都认为你的MySQL引擎是InnoDB）。没有SQL错误，恐怕没有人可以帮助你。