PHP 用户名逻辑不起作用

Question

我遇到的错误:无论我尝试输入有效的用户名还是密码，它总是显示“抱歉，用户名已存在”。我的逻辑有问题吗，特别是这个代码块:

 $q = $conn->query("SELECT uname FROM Student WHERE uname= $aRegUsr ");
    $stmt = $conn->prepare($q);
    $stmt->fetch($q);
    if($stmt > 0){ #if there are 1 or more users with enter username, deny.

dbcred.php

<?php
$host = 'localhost';
$user = 'root';
$pass = '';
$db = 'databasename';

?>

PHP:

<?php
#Login Details
require_once('dbcred.php');
$conn = new PDO("mysql:host=$host;dbname=$db", $user, $pass);

$aRegUsr = $_POST['regduser']; 

#Check for Existing User
$q = $conn->query("SELECT uname FROM Student WHERE uname= $aRegUsr ");
$stmt = $conn->prepare($q);
$stmt->fetch($q);
if($stmt > 0){ #if find that username in use, deny ability to register, else{ALLOW}
echo "Sorry, username already exists";
}
else{
echo "Success";
}

?>

HTML

<form action="inc/check_regUsr.php" method="post" id="userLogon">
    <div class="field required">
        Username: <input type="text" name="regduser" tabindex="1" /><br />
        </div>
        <div class="field required">
        Password: <input type="password" name="regdpass" tabindex="2" /><br />
        </div>
        <input type="submit" name="submitUser" />
</form>

数据库表:学生

Answer 1

您是否正在尝试创建用户注册或用户身份验证脚本？我不认为您的实际目的已明确概述以帮助提供有效的解决方案。

“我是否尝试输入有效的用户名或密码”表示用户身份验证，而“if($stmt > 0){ #if There are 1 or more users with Enter username, Deny.”表示用户注册。

编辑:根据您更新的要求，用户身份验证可能会变得有点复杂，具体取决于您的部署的安全性。下面包含基于您的基本代码的示例代码。请注意，您可能需要找到一个预先构建的脚本并根据您的需要对其进行修改，因为这可能会更好地满足您的目的。

<?php #Login Details 
require_once('dbcred.php'); 

#Variables
$error = '';
$auth = 0;
$aRegUsr = $_POST['regduser'];   #Check for Existing User 
$aRegPass = $_POST['regdpass'];   #Check for Existing User 

#DB stuff
$conn = new PDO("mysql:host=$host;dbname=$db", $user, $pass); 
$sql = "SELECT uname,password FROM Student WHERE uname= $aRegUsr ";

#Dataset stuff
foreach($conn->query($sql) as $row){ 
    if($row['password'] != $aRegPass){
        $error = 'Invalid Password';
        $auth = 0; #If $auth = 0 then user is logged in
    }else{
        $auth = 1; #If $auth = 1 then user is logged in
    }
}
$conn = null;
?>