在PHP包含頁面的jQuery驗證後無法驗證PHP

Question

有人可以幫我弄清楚如何讓我的表單在jQuery驗證後提交給php。

我找到了一個教程「構建一個整潔的HTML5動力聯繫表單」，並且如果我構建表單以便它是一個完整的頁面，並且頭部和身體驗證。但是我的網站被設計頁眉和頁腳是所有頁面和我一樣包括中央例如不同的內容：

[HTTP：//mysite/index.php頁=接觸]

這是JS腳本：

`$(function(){ 
     $("#refreshimg").click(function(){ 
      $.post('newsession.php'); 
      $("#captchaimage").load('image_req.php'); 
      return false; 
     }); 

     $("#contact_form").validate({ 
      rules: { 
        name: { 
        required: true, 
        minlength: 2 
        }, 
        //telephone, email & message rules removed 

       captcha: { 
        required: true, 
        remote: "process.php" 
       } 
      }, 
      //messages removed 

      errorContainer: $('#errors'), 
      errorLabelContainer: $('#errors ul'), 
      wrapper: 'li', 

      var dataString = 'name='+ name + '&email=' + email + '&telephone=' + telephone + '&message=' + message; 
       //alert (dataString);return false; 
       submitHandler: function(form) { 
       $(form).ajaxSubmit({ 
       type: "POST", 
       url: "pages/processC.php", 
       data: dataString, 
       success: function() { 
        $('#contactForm').html("<div id='message'></div>"); 
        $('#message').html("<h2>Contact Form Submitted!</h2>") 
        .append("<p>We will be in touch soon.</p>") 
        .hide() 
        .fadeIn(1500, function() { 
        $('#message').append("<img id='checkmark' src='images/check.png' />"); 
        }); 
       } 
       }); 
       } 
      return false; 

      onkeyup: false 
     }); ` 

The tutorial I followed came with the following submitHandler: 

` /*submitHandler: function() { 
      alert("Correct captcha!"); 
     }, 
     success: function(label) { 
      label.addClass("valid").text("Valid captcha!") 
     },*/` 

其中我註釋掉，當它是一個完整的頁面表單提交。

我在另一個教程中找到了上面的ajaxSubmit，但是當我點擊提交時，它會導航到我的索引頁面。

這是我的PHP文件：

` >

<?php 
if(isset($_POST)){ 
    //form validation vars 
    $formok = true; 
    $p_errors = array(); 

    //sumbission data 
    $ipaddress = $_SERVER['REMOTE_ADDR']; 
    $date = date('d/m/Y'); 
    $time = date('H:i:s'); 

    //form data 
    $name = filter_var($_POST['name'], FILTER_SANITIZE_STRING); 
    $email = filter_var($_POST['email'], FILTER_SANITIZE_EMAIL); 
    $telephone = filter_var($_POST['telephone'], FILTER_SANITIZE_NUMBER_INT); 
    $message = filter_var($_POST['message'], FILTER_SANITIZE_STRING); 
    $captcha = filter_var($_POST['captcha'], FILTER_SANITIZE_STRING); 

    //validate form data 

    //validate name is not empty 
    if(empty($name)){ 
     $formok = false; 
     $errors[] = "You have not entered a name"; 
    } 

    //validate telephone is numbers 
    if(empty($telephone)){ 
     $formok = true; 
    } 

    //validate email address is not empty 
    if(empty($email)){ 
     $formok = false; 
     $errors[] = "You have not entered an email address"; 
    //validate email address is valid 
    } 

    //validate message is not empty 
    if(empty($message)){ 
     $formok = false; 
     $errors[] = "You have not entered a message"; 
    } 

    //validate captcha is not empty 
    session_start(); 
    if(empty($captcha)){ 
     $formok = false; 
     $errors[] = "Please answer captcha question"; 
    } 
    elseif(strtoupper($_POST['captcha']) == $_SESSION['captcha_id']) 
    { 
    //Do your stuff 
    unset ($_SESSION["captcha_id"]); 
    } 
    else 
    { 
    $formok = false; 
    $errors[] = "Wrong code entered"; 
    } 

    //send email if all is ok --REMOVED 


    //what we need to return back to our form 
    $returndata = array(
     'posted_form_data' => array(
      'name' => $name, 
      'email' => $email, 
      'telephone' => $telephone, 
      'message' => $message 
     ), 
     'form_ok' => $formok, 
     'errors' => $errors 
    ); 

    //if this is not an ajax request 
    if(empty($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) !== 'xmlhttprequest'){ 
     //set session variables 
     session_start(); 
     $_SESSION['cf_returndata'] = $returndata; 

     //redirect back to form 
     header('location: ' . $_SERVER['HTTP_REFERER']); 
    } 
}` 

我希望我沒有把太多的代碼，但我的PHP和jQuery的知識是非常有限的，但我真的需要解決？這樣所以任何建議將非常感激。

由於提前， 詹姆斯

Answer 1

不要試圖用手構建您的文章的身體。使用您的方法會引入很多編碼問題。

相反的：

var dataString = 'name='+ name + '&email=' + email + '&telephone=' + telephone + '&message=' + message; 

只是

var dataObj = {"name":name, "email":email, "telephone":telephone}; 

然後在你的Ajax呼叫改變

data: dataString, 

到

data: dataObj; 

這將避開很多角落案例和編碼問題。