Question

所以我有一个注册，登录系统，一旦用户登录，他们就在他们的帐户中。从他们的帐户，他们可以填写其他页面上的其他详细信息，即个人详细信息表一旦他们填写了这些细节，它应该将它们返回到他们的帐户，但显然他们的详细信息应该在个人表中的mysql数据库中，但链接到他们在users表中的id（存储注册详细信息）我有点卡在我的个人形式的代码上，这确保了特定用户的id是相同的。任何帮助都会很棒

例如，我在两个表中都有'id'。我做了这个汽车公司。我也有一个名为user_id的唯一字段，用于链接表。

我个人的PHP代码：

include 'dbc.php';

$err = array();

if (!empty($_POST['doPersonal']) && $_POST['doPersonal'] == 'Submit') 
{ 
/******************* Filtering/Sanitizing Input *****************************
This code filters harmful script code and escapes data of all POST data
from the user submitted form.
*****************************************************************/
 foreach($_POST as $key => $value) {
$data[$key] = filter($value);
}







$host  = $_SERVER['HTTP_HOST'];
$host_upper = strtoupper($host);
$path   = rtrim(dirname($_SERVER['PHP_SELF']), '/\\');

if(empty($err)) {

$sql_insert = "INSERT into `personal`
        (`sex`,`aux_citizen`,`birth_place`,`birth_country`,`children`
        )
        VALUES
          ('$data[sex]','$data[aux_citizen]','$data[birth_place]',
'$data[birth_country]','$data[children]'
        )
        ";

mysql_query($sql_insert,$link) or die("Insertion Failed:" . mysql_error());




header("Location: thankyou.php");  
exit();

 } 
}

部分用户（注册详情）php代码：

<?php 

include 'dbc.php';

$err = array();

if(@$_POST['doRegister'] == 'Register') 
 { 
   foreach($_POST as $key => $value) {
$data[$key] = filter($value);
 }


$user_ip = $_SERVER['REMOTE_ADDR'];

$sha1pass = PwdHash($data['pwd']);

$host  = $_SERVER['HTTP_HOST'];
$host_upper = strtoupper($host);
$path   = rtrim(dirname($_SERVER['PHP_SELF']), '/\\');

$usr_email = $data['usr_email'];
$user_name = $data['user_name'];



$birthday = date("Y-m-d", mktime(0,0,0,$dt,$mo,$yr));
if(empty($err)) {

$sql_insert = "INSERT into `users`
        (`full_name`,`user_email`,`pwd`,`nationality`,`department`
,`birthday`,`date`,`users_ip`,`activation_code`,`user_name`
        )
        VALUES
        ('$data[full_name]','$usr_email','$sha1pass'
,'$data[nationality]','$data[department]','$birthday'
        ,now(),'$user_ip','$activ_code','$user_name'
        )
        ";

mysql_query($sql_insert,$link) or die("Insertion Failed:" . mysql_error());
$user_id = mysql_insert_id($link);  
$md5_id = md5($user_id);
mysql_query("update users set md5_id='$md5_id' where id='$user_id'");
//  echo "<h3>Thank You</h3> We received your submission.";
?>

Answer 1

究竟是什么问题？将ID保存在PHP中，或将ID字段从个人链接到用户的ID字段？

如何保持相同的用户配置文件

1 个答案: