我使用以下脚本在mysql中创建了一个表：

CREATE TABLE IF NOT EXISTS users_x_activities(
  id int NOT NULL auto_increment, 
  id_user int unsigned NOT NULL,
  id_attivita int unsigned NOT NULL,
  PRIMARY KEY (id),
  FOREIGN KEY (id_user) REFERENCES utente(id),
  FOREIGN KEY (id_attivita) REFERENCES attivita(id)
  ) ENGINE = INNODB;

当我从phpMyAdmin导出创建的表时，我获得了以下脚本

CREATE TABLE IF NOT EXISTS `users_x_activities` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `id_user` int(10) unsigned NOT NULL,
  `id_attivita` int(10) unsigned NOT NULL,
  PRIMARY KEY (`id`),
  KEY `id_user` (`id_user`),
  KEY `id_attivita` (`id_attivita`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=1 ;

所以问题是：我的外键约束在哪里？KEY引用FK吗？看起来utente和attivita这两个表在新生成的脚本中不再被引用。我哪里做错了？

编辑

在phpMyAdmin中，配置表的导出时，我发现了选项“Display Foreign Key Relationship”（显示外键关系）。如果我标记了此选项，我还将在脚本中获得此代码

--
-- RELATIONS FOR TABLE `users_x_activity`:
--   `id_user`
--       `utente` -> `id`
--   `id_attivita`
--       `attivita` -> `id`
--

--
-- Constraints for dumped tables
--

--
-- Constraints for table `users_x_activity`
--
ALTER TABLE `users_x_activity`
  ADD CONSTRAINT `users_x_activities_ibfk_1` FOREIGN KEY (`id_user`) REFERENCES `utente` (`id`),
  ADD CONSTRAINT `users_x_activities_ibfk_2` FOREIGN KEY (`id_attivita`) REFERENCES `attivita` (`id`);

这意味着，如果我添加选项“显示外键关系”，我也获得FK约束？在其他情况下不是？