php - POST undefined variable ，但 REQUEST 有效

Question

我收到以下所有变量的错误消息 undefined variable :

if(isset($_POST['nome'])){
    $name = $_POST['nome'];
}

if(isset($_POST['email'])){
    $email = $_POST['email']; 
}


if(isset($_POST['password'])){
    $password = $_POST['password']; 
}

如果我使用 REQUEST 而不是 POST 它可以工作，但我读到用户表单最好使用 POST。

这是我的整个 PHP 代码:

<?php
 define('DB_HOST', 'localhost');
 define('DB_NAME', 'class_rate');
 define('DB_USER', 'root');
 define('DB_PASSWORD', 'vB42lL&69_r');

 $con = mysqli_connect(DB_HOST, DB_USER);
    if(!$con){
        die("Databese Connection Failed" . mysqli_error($con));
    }
 $db = mysqli_select_db($con, DB_NAME);
    if(!$db){
        die("Databese Selection Failed" . mysqli_error($con));
    }



   function NewUser($con, $db){

            if(isset($_POST['nome'])){
        $name = $_POST['nome'];
    }

    if(isset($_POST['email'])){
        $email = $_POST['email']; 
    }


    if(isset($_POST['password'])){
       $password = $_POST['password']; 
    }

    $query = "INSERT INTO users (nome_user,email_user,passw_user) VALUES ('$name','$email','$password')"; 
     $data = mysqli_query($con, $query)or die(mysqli_error($con)); 
    if($data) { 
        echo "YOUR REGISTRATION IS COMPLETED..."; 
    }
 }

function SignUp($con,$db){
     if (!empty($_POST['email'])) {
      $query = mysqli_query("SELECT * FROM users WHERE email_user = '$_POST[email]' AND passw_user = '$_POST[password]'") or die(mysqli_error($con));

       if(!$row = mysqli_fetch_array($query) or die(mysqli_error($con)) ){
            NewUser($con,$db);
       }
}

   else{
      echo "Email already registered!";
    }
}

   if($_SERVER['REQUEST_METHOD'] == "POST"){
      SignUp($con,$db);
   }


 ?>

我不知道我的问题是否出在我的表单 HTML 代码上，所以我也将它包括在这里:

 <form action="cadastro.php" method="post">
        <label><b>Nome:</b></label>
        <div>
            <input type="text" placeholder="Nome" id = "nome"name="nome" required>              
        </div>
        <label><b>Email:</b></label>
        <div>
            <input type="email" placeholder="Email" id = "email "name="email" required>
        </div>
        <label><b>Confirmar Email:</b></label>
        <div>
            <input type="email" placeholder="Confirmar Email" id="confirmar_email" name="confirmar_email" required>     
        </div>
        <label><b>Universidade:</b></label>
        <div>               
            <input type="text" placeholder="Universidade" id="universidade" name="universidade" required>
        </div>
        <label><b>Curso:</b></label>
        <div>
            <input type="text" placeholder="Curso" id="curso" name="curso" required>
        </div>
        <label><b>Senha:</b></label>
        <div>
            <input type="password" placeholder="Senha" id="password" name="password" required>
        </div>
        <label><b>Confirmar Senha:</b></label>
        <div>
            <input type="password" placeholder="Confirmar Senha" id="confirmar_password" name="confirmar_password" required>
        </div>
        <input type="checkbox"> Ao apertar na caixa voce confirma que leu e conconrda com os <a href="#">Termos e Condicoes</a>.
        <div class="botoes">
            <button name = "sub" id = "sub" type="submit" class="signupbtn">Confirmar</button>
            <button name = "cancel" id = "cancel" type="button" class="cancelbtn">Cancelar</button>
        </div>
  </form>

Answer 1

正如您所说的，您的 PHP 脚本完全容易受到所有类型的攻击。请使用 Php PDO & 不要避免编码安全。关于您的问题:

改变这个

<button type='submit' ></button>

到

 <input type='submit' value='submit' />

确保所有这些都在表单标签内。当使用某些 java 脚本时首选按钮，但在您的代码中我没有看到任何 java 脚本，因此请避免使用它。希望对您有所帮助。