表格的問題

Question

我正在嘗試使用js和php做一個表單，所以，當我點擊按鈕時，它會一直保持在屏幕「發送電子郵件...」並且永遠不會發送，我檢查了控制檯，它說，錯誤是在我的PHP文件（404未找到錯誤）......如果有人能幫助我，我留下來使用的代碼如下：

HTML：

<div class="contact-form"> 
    <h3>Contate-nos</h3> 
      <form id="main-contact-form" name="contact-form" method="post" action="email.php"> 
       <div class="form-group"> 
        <input type="text" name="name" class="form-control" id="name" placeholder="Nome"> 
       </div> 
       <div class="form-group"> 
        <input type="email" name="email" class="form-control" id="email" placeholder="Email"> 
       </div> 
       <div class="form-group"> 
        <input type="text" name="subject" class="form-control" id="subject" placeholder="Assunto"> 
       </div> 
       <div class="form-group"> 
        <textarea name="message" class="form-control" rows="8" id="message" placeholder="Message"></textarea> 
       </div> 
       <button type="submit" name="submit" class="btn btn-primary">Send Message</button> 
     </form> 
</div> 

JS：

<script> 
     var form = $('#main-contact-form'); 
     form.submit(function(event){ 
      event.preventDefault(); 
     var form_status = $('<div class="form_status"></div>'); 
      var data = { 
       name: $('#name').val(), 
       email: $('#email').val(), 
       subject: $('#subject').val(), 
       message: $('#message').val() 
      }; 
      $.ajax({ 
       url: $(this).attr('action'), 
       type: "POST", 
       dataType: "json", 
       data: {'data': data}, 
       beforeSend: function(){ 
        form.prepend(form_status.html('<p><i class="fa fa-spinner fa-spin"></i> Sending email...</p>').fadeIn()); 
       }, 
      success: function (data) { 
       if(data == 1){ 
        form_status.html('<p class="text-success">Sucessful!</p>').delay(3000).fadeOut(); 
       } 
       else{ 
        form_status.html('<p class="text-error">Sorry, try again later!</p>').delay(3000).fadeOut(); 
       } 
      }, 
      error: function (xhr, status, error) { 
       console.log(error); 
      } 
     }); 
    }); 
    </script> 

PHP：

<?php 
$data = $_POST['data']; 
$name  = $data['name']; 
$from  = $data['email']; 
$subject = $data['subject']; 
$message = $data['message']; 
$to   = 'user@email.com'; 
$headers = "MIME-Version: 1.0"; 
$headers .= "Content-type: text/plain; charset=iso-8859-1"; 
$headers .= "From: {$name} <{$from}>"; 
$headers .= "Reply-To: <{$from}>"; 
$headers .= "Subject: {$subject}"; 
$headers .= "X-Mailer: PHP/".phpversion(); 
if(mail($to, $subject, $message, $headers)){ 
    echo 1; 
} 
else{ 
    echo 0; 
} 

Answer 1

你一個AJAX網址錯了，你應該這樣定義

$.ajax({ 
    url: 'frmSelf.php', 
    data: $("#frmSelf").serialize(), 
    dataType: 'json', 
    type : 'post', 
    success : function(returnData) { 
     console.log(returnData); 
    } 
});