Question

我收到以下所有错误消息的未定义变量：

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，但我认为最好将POST用于用户表单。

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与表单的HTML代码有关，所以我也将其包含在这里：

 <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

正如你所说的所有php脚本所说的那样，对于所有类型的攻击都是完全容易受到攻击的。请使用Php PDO＆amp;不要避免编码安全性。关于你的问题：

更改此

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

要

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

确保所有这些内部形式标记。当使用某些java脚本时，按钮是首选，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。希望它有所帮助。

POST未定义的变量，但REQUEST有效

1 个答案: