2013-05-20 22 views
-2

不能看出爲什麼這個聯繫表單不工作任何指針將是偉大的...想通過ajax發送表單,並串行化數據發送到..只是沒有看起來是工作though..hope一些人可以幫PHP簡單的聯繫表單不工作不確定爲什麼

HTML

<form method="post" action="" id="contact_form"> 
    <div class="row"> 
       <div class="large-6 columns"> 
        <label for="contact_name">Your Name</label> 
        <input type="text" placeholder="" id="contact_name" name="contact_name" class="required"> 
       </div> 
       <div class="large-6 columns"> 
        <label for="contact_email">Your Email</label> 
        <input type="text" placeholder="" id="contact_email" name="contact_email" class="required"> 
       </div> 
      </div> 
      <div class="row"> 
       <div class="large-6 columns"> 
        <label for="contact_company">Company Name or Organization</label> 
        <input type="text" placeholder="" id="contact_company" name="contact_company" class="required"> 
       </div> 
       <div class="large-6 columns"> 
        <label for="contact_phone">Phone Number</label> 
        <input type="text" placeholder="" id="contact_phone" name="contact_phone" class="required"> 
       </div> 
      </div> 
      <div class="row"> 
       <div class="large-12 columns"> 
        <label for="contact_message">Your Message</label> 
        <textarea rows="4" placeholder="" id="contact_message" name="contact_message" class="required"></textarea> 
       </div> 
      </div> 
      <div class="row"> 
       <div class="large-12 columns"> 
        <input type="submit" name="contact_submit" id="contact_submit" value="SEND MESSAGE" class="button"> 
       </div> 
      </div> 

     </form> 

     <p class="success" style="display:none">Your message has been 
       sent successfully.</p> 


Mail.php

<?php 

$name = $POST['contact_name'] ; 
$email = $POST['contact_email'] ; 
$company = $_POST['contact_company'] ; 
$number = $_POST['contact_phone'] ; 
$message = $POST['contact_message'] ; 

mail("[email protected]", $name, $company, $number, $message, "From:" . $email); 
    ?> 

JS文件

$(document).ready(function() { 

    $('#contact_form').validate({ 

    submitHandler: function(form) { 
     //do submit 

      var dataString = $("this").serialize(); 
      //alert (dataString);return false; 
      $.ajax({ 
      type: "POST", 
      url: "php/mail.php", 
      data: dataString, 
      success: function() { 
      $('.success').fadeIn(1000); 
       $("input[type=text], textarea").val(""); 
       $('.success').fadeToggle(1000); 
     } 
     }); 
      return false; 
    } 
}); 
}); 
+1

變化'$ POST'到'$ _POST'並請解釋一下你已經嘗試解決它 – kero

+0

嗨,這首先要求進行基本的調試。究竟什麼地方不起作用?測試結果是什麼在JavaScript和PHP中顯示出來 - 是否有過這樣的要求?數據是否傳輸過? –

+0

我現在正在工作,但它似乎沒有發送數據,並被放入我的垃圾郵件..怪異.. – AC88

回答

3

您犯了一些錯誤。使用下面的代碼:

<?php 
$name = $_POST['contact_name'] ; 
$email = $_POST['contact_email'] ; 
$company = $_POST['contact_company'] ; 
$number = $_POST['contact_phone'] ; 
$message = $_POST['contact_message'] ; 

//modify the mail function 
mail("[email protected]", $name.$company, $message, "From:" . $email); 
?> 

而且在JS改變如下:

success: function(returnData) { 
     $('.success').fadeIn(1000); 
     $("input[type=text], textarea").val(""); 
     //$('.success').fadeToggle(1000); 

    } 
相關問題