Ajax沒有得到來自PHP端的響應

Question

在我的註冊表單中使用ajax + php。有2個驗證：JS端前端和PHP端後端。在PHP端創建了一個名爲response的特殊函數：它將PHP端錯誤作爲JSON數據發送。

問題是我無法從PHP方面得到任何迴應。

在firebug中分析過的頁面：收到錯誤消息responseData is null。 （responseData = jQuery.parseJSON(data)）

JS部分看起來像

//check the form is not currently submitting 
    if ($(this).data('formstatus') !== 'submitting') { 

var form = $(this), 
    formData = form.serialize() + '&formID=' + form.attr('id'), 
    formUrl = form.attr('action'), 
    formMethod = form.attr('method'); 



//add status data to form 
form.data('formstatus', 'submitting'); 

if (validate()) { 
    //send data to server for validation 
    $.ajax({ 
     url: formUrl, 
     type: formMethod, 
     data: formData, 
     success: function (data) { 

      //setup variables 
      var responseData = jQuery.parseJSON(data), 
      cl, text; 

      //response conditional 
      switch (responseData.status) { 
      case 'error': 
      cl = 'error'; 
      text = responseData.message; 
      break; 
      case 'success': 
      cl = 'success'; 
      text = 'Qeydiyyat uğurla başa çatdı'; 
      break; 
      } 


      $.notifyBar({ 
      cls: cl, 
      html: text 
      }); 

     } 
    }); 

} 
form.data('formstatus', 'idle'); 


} 

這裏是PHP部分

<?php 
require '../common.php'; 

function checkIfEmailExists($email, $stmt) 
{ 
     if ($stmt = $db->prepare("SELECT id FROM TABLE WHERE email=? LIMIT 1")) { 
       $stmt->bind_param("s", $email); 
       $stmt->execute(); 
       $stmt->bind_result($count); 
       $stmt->close(); 
     } 

     return ($count > 0 ? true : false); 
} 


if ($_POST['formID'] == 'signup_form') { 
     // Setting vars 
     $lname  = $_POST['lname']; 
     $fname  = $_POST['fname']; 
     $mname  = $_POST['mname']; 
     $email  = $_POST['email']; 
     $pass   = $_POST['pass']; 
     $confirm_pass = $_POST['confirm_pass']; 

     //===================== 
     //Server side validation >> 


     //First name, middle name, last name check >> 
     if (!$lname) { 
       response('error', 'Familiyanı daxil edin'); 
     } 
     if (!$fname) { 
       response('error', 'Adı daxil edin'); 
     } 
     if (!$mname) { 
       response('error', 'Atanızın adını daxil edin'); 
     } 
     //<< 

     //Pass check >> 
     if (strlen($pass) > 2) { 
       if ($pass == $confirm_pass) { 
         return true; 
       } else { 
         response('error', 'Şifrənin təkrarlanmasında səhv'); 
       } 
     } else { 
       response('error', 'Şifrədə simvolların sayı 4-dən çox olmalıdır'); 
     } 

     //<< 


     //email validation >> 
     if (filter_var($email, FILTER_VALIDATE_EMAIL)) { 
       if (!checkIfEmailExists($email, $stmt)) { 
         return true; 
       } else { 
         response('error', 'Bu ünvanla qeydiyyata alınmış başqa istifadəçi var.'); 
       } 
     } else { 
       response('error', 'Email ünvanını düzgün daxil edin'); 
     } 

     //<< 

     // Create statement object 
     $stmt = $db->stmt_init(); 

     // Create a prepared statement 
     if ($stmt->prepare("INSERT INTO `users` (`fname`, `mname`, `lname`, `email`, `pass`, `reg_dt`) VALUES (?, ?, ?, ?, ?, NOW())")) { 
       // Binding vars 

       $rc = $stmt->bind_param('sssss', $fname, $lname, $mname, $email, $pass) or die('bind_param() failed: ' . htmlspecialchars($stmt->error)); 

       // Execute query 
       $rc = $stmt->execute(); 
       if ($rc) { 
         response('success', 'Qeydiyyat uğurla başa çatdı'); 
       } else { 
         response('error', htmlspecialchars($stmt->error)); 
       } 


       // Close statement object 
       $stmt->close(); 

     } else { 
       response('error', htmlspecialchars($dv->error)); 
     } 



} 
else {response('error', 'Qeydiyyatda problem');} 

     //return json response 
     function response($status, $message) 
     { 
       $data = array(
         'status' => $status, 
         'message' => $message 
       ); 
       echo json_encode($data); 
       die(); 
     } 
?> 

Answer 1

您需要添加

dataType: "json", 

在$。 Ajax方法。