Php 注册未验证密码的用户脚本

好的，我的用户注册脚本有点问题，出于某种原因，它没有检查表单字段pass1和pass2是否相同，尽管它在代码中，有什么想法吗？我将把注册代码完整地放在下面，供您审阅。我对PHP非常陌生，因此非常感谢您的任何建议，谢谢您的帮助。：-）

注册脚本

<?php
ob_start(); // Start output buffering

function isLoggedIn()
{
    if(isset($_SESSION['valid']) && $_SESSION['valid'])
        return true;
    return false;
}


session_start();
//if the user has not logged in
if(!isLoggedIn())
{
    header('Location: ../index.php');

    die();
}


?>
<?php require_once('../Connections/PropSuite.php'); ?>

<?php

error_reporting(E_ALL & ~E_NOTICE);
ini_set('display_errors', TRUE);
ini_set('display_startup_errors', TRUE);

if (!function_exists("GetSQLValueString")) {
function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "") 
{
  if (PHP_VERSION < 6) {
    $theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue;
  }

  $theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($theValue) : mysql_escape_string($theValue);

  switch ($theType) {
    case "text":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;    
    case "long":
    case "int":
      $theValue = ($theValue != "") ? intval($theValue) : "NULL";
      break;
    case "double":
      $theValue = ($theValue != "") ? doubleval($theValue) : "NULL";
      break;
    case "date":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;
    case "defined":
      $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue;
      break;
  }
  return $theValue;
}
}

$colname_user = "-1";
if (isset($_SESSION['username'])) {
  $colname_user = $_SESSION['username'];
}
mysql_select_db($database_Takeaway, $Takeaway);
$query_user = sprintf("SELECT type FROM admin_users WHERE username = %s", GetSQLValueString($colname_user, "text"));
$user = mysql_query($query_user, $Takeaway) or die(mysql_error());
$row_user = mysql_fetch_assoc($user);
$totalRows_user = mysql_num_rows($user);






  // Username available.


//form begins
 $user_type = $row_user['type'];

    if ($user_type === 'admin-full') 
    {
//retrieve our data from POST
$name = $_POST['name'];
$username = $_POST['username'];
$email = $_POST['email'];
$pass1 = $_POST['pass1'];
$pass2 = $_POST['pass2'];
if($pass1 != $pass2)
    header('Location: register-admin.php?pw=notmatched');
if(strlen($username) > 30)
    header('Location: register-admin.php?username=toolong');
if(strlen($name) > 40)
    header('Location: register-admin.php?name=toolong');

$hash = hash('sha256', $pass1);

//creates a 3 character sequence
function createSalt()
{
    $string = md5(uniqid(rand(), true));
    return substr($string, 0, 3);
}
$salt = createSalt();
$hash = hash('sha256', $salt . $hash);


mysql_select_db($database_Takeaway, $Takeaway);
//sanitize username
$username = mysql_real_escape_string($username);

$q = "SELECT id, username FROM admin_users WHERE username = '$username'";
$results = @mysql_query($q);
if(mysql_num_rows($results) > 0)
{
  // Username exists.
 header('Location: register-admin.php?username-taken');
}
else
{


$query = "INSERT INTO admin_users ( type, email, name, username, password, salt )
        VALUES ( 'admin-full', '$email', '$name', '$username' , '$hash' , '$salt' );";
mysql_query($query);
mysql_close();
header('Location: ../main?NewUserAdded');
}

mysql_free_result($user);
    } 
    else {
?>
You are not authorised to do that!
<?php } ?>

放一个
die（）在
标题之后（'位置：register admin.php？pw=notmatched'）
像这样：

if($pass1 != $pass2) {
    header('Location: register-admin.php?pw=notmatched');
    die();
}
if(strlen($username) > 30) {
    header('Location: register-admin.php?username=toolong');
    die();
}
if(strlen($name) > 40) {
    header('Location: register-admin.php?name=toolong');
    die();
}

似乎您希望用户名值在查询数据库以获取用户信息时处于会话中。我不明白如果你只是发布一个表单，用户名在会话中会是什么样子。你的条件

if ($user_type === 'admin-full')

似乎它永远不会计算为真，因为您还没有从post数组中提取用户名并对其进行计算

注意，您还应该使用
mysqli
或
PDO
而不是
mysql.*
函数，因为这些函数在PHP中已被弃用。
Hi，是的，这是一个管理员用户添加另一个管理员用户，因此我需要只向拥有完整管理员帐户的人授予访问该页面的权限，该
GetSQLValueString
功能绝对可怕。除了混淆你的不计后果的漠视之外，这还有什么作用？
if ($user_type === 'admin-full')