在表'用户'中注册我的第一个用户时,id与链接表中的user_id相同,为1(语言)。然而,当我注册的其他用户(ID2中的用户)的语言USER_ID仍为1请参见SQL:外键无法正确链接
CREATE TABLE `language` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`user_id` int(11) NOT NULL,
`native` varchar(30) NOT NULL,
`other` varchar(30) NOT NULL,
`other_list` varchar(9) NOT NULL,
`other_read` varchar(9) NOT NULL,
`other_spokint` varchar(9) NOT NULL,
`other_spokprod` varchar(9) NOT NULL,
`other_writ` varchar(9) NOT NULL,
PRIMARY KEY (`id`),
KEY `user_id` (`user_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=9 ;
CREATE TABLE IF NOT EXISTS `users` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`md5_id` varchar(200) NOT NULL,
`full_name` tinytext CHARACTER SET latin1 COLLATE latin1_general_ci NOT NULL,
`user_name` varchar(10) NOT NULL,
`user_email` varchar(30) NOT NULL,
`user_level` tinyint(4) NOT NULL DEFAULT '1',
`pwd` varchar(220) NOT NULL,
`nationality` varchar(30) NOT NULL,
`department` varchar(20) NOT NULL,
`birthday` date NOT NULL,
`date` date NOT NULL DEFAULT '0000-00-00',
`users_ip` varchar(200) NOT NULL,
`activation_code` int(10) NOT NULL DEFAULT '0',
`banned` int(1) NOT NULL,
`ckey` varchar(200) NOT NULL,
`ctime` varchar(220) NOT NULL,
`approved` int(1) NOT NULL DEFAULT '1',
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ;
这是我的PHP代码:
if(empty($_SESSION['$user_id'])) { // user not logged in; redirect to somewhere else }
if (!empty($_POST['doLanguage']) && $_POST['doLanguage'] == 'Submit')
{
$result = mysql_query("SELECT `id` FROM users WHERE `banned` = '0'") or
die (mysql_error());
list($id) = mysql_fetch_row($result);
session_start();
$_SESSION['user_id']= $id;
sql_insert = "INSERT into `language`
(`user_id`,`native`,`other`,`other_list`,`other_read`, `other_spokint`
,`other_spokprod`,`other_writ` )
VALUES
('$id','$native','$other','$other_list','$other_read','$other_spokint',
'$other_spokprod','$other_writ') ";
mysql_query($sql_insert,$link) or die("Insertion Failed:" . mysql_error());
}
header("Location: myaccount.php?id=' . $_SESSION[user_id] .'");
exit();
希望得到任何帮助!
$ id在哪里填充?或者是这条线向后写:“$ _SESSION ['user_id'] = $ id;” – Yamikuronue 2012-03-28 16:06:24
@Yamikuronue - 它来自与获取结果集一起工作的'list'构造。 – 2012-03-28 16:32:01
@Yamikuronue我试图写它的另一种方式,但仍然有user_id 1的问题,当它应该是2.thanks – user1296762 2012-03-28 16:38:24