Php Jquery表单插件未提交到流程页面

我有一个使用Jquery表单插件的基本联系人表单。我不能让它提交使用插件。它只有在我向process.php添加一个php include时才会提交

Javascript:

 $(document).ready(function() { 
     $("#contact").validate({
         submitHandler: function(form) {
             $(form).ajaxSubmit({
                 url:"process.php",
                 type:"post",
             });
         }
     });
 });

这是我的联系方式：

 <form id="contact" method="post" name="validation" action="contact.php" onsubmit="return validateForm();">
     <fieldset> 

         <label for="name">Name</label>
         <input type="text" name="name" placeholder="Full Name" title="Enter your name" class="required">

         <label for="email">E-mail</label>
         <input type="email" name="email" placeholder="yourname@domain.com" title="Enter your e-mail address" class="required email">

         <label for="phone">Phone</label>
         <input type="tel" name="phone" placeholder="ex. (555) 555-5555">

         <label for="message">Question/Comment</label>
         <textarea name="message"></textarea>

         <input type="submit" name="submitted" class="button" id="submit" value="Send Message"     />

     </fieldset>
 </form>

名称
电子邮件
电话
问题/评论

Process.php

   <?php
   function GetHeaders()
   {
    $headers  = 'MIME-Version: 1.0' . "\r\n";
    $headers .= 'Content-type: text/html; charset=iso-8859-1' . "\r\n";
    // Additional headers
    $headers .= 'From: Company Name<info@companyname.com>' . "\r\n";
    return $headers;
   }
   // Get Data  
   $name = strip_tags($_POST['name']);
   $email = strip_tags($_POST['email']);
   $phone = strip_tags($_POST['phone']);
   $message = strip_tags($_POST['message']);
    // Send Message
    $headers = GetHeaders();
    $intro = "\"Thank you for contacting Company Name. We are very interested in assessing your situation and will be in touch as soon possible.\" <br />
    <br/>
      Best Regards,<br/>
    <br/>
      Company<br/>
    ";
 mail($email, "RE: Contact Form Submission", $intro, $headers);      
 ?>

action="contact.php"

---

感谢您提前提供的帮助。

您需要从表单属性中删除

onsubmit=“return validateForm（）；

。您已经在使用

$（“#contact”）进行验证。验证（{

在查询就绪函数中

另外，您的表单似乎提交到contact.php，而不是process.php

   <?php
   function GetHeaders()
   {
    $headers  = 'MIME-Version: 1.0' . "\r\n";
    $headers .= 'Content-type: text/html; charset=iso-8859-1' . "\r\n";
    // Additional headers
    $headers .= 'From: Company Name<info@companyname.com>' . "\r\n";
    return $headers;
   }
   // Get Data  
   $name = strip_tags($_POST['name']);
   $email = strip_tags($_POST['email']);
   $phone = strip_tags($_POST['phone']);
   $message = strip_tags($_POST['message']);
    // Send Message
    $headers = GetHeaders();
    $intro = "\"Thank you for contacting Company Name. We are very interested in assessing your situation and will be in touch as soon possible.\" <br />
    <br/>
      Best Regards,<br/>
    <br/>
      Company<br/>
    ";
 mail($email, "RE: Contact Form Submission", $intro, $headers);      
 ?>

action="contact.php"

---

process.php

的内容是什么？我在上面添加了它，但在使用include（“process.php”）时，它似乎还没有到达process.php；我可以让一切正常工作，但我真的很想弄清楚为什么表单插件不起作用。我还在html

action='contact.php'

和ajax

url:'process.php'

中使用jquery 1.8.0，这让我很困惑Hanks，hsalama——我想插件会覆盖表单上的操作，即contact.php？我不想让用户失望o提交表单后，请转到process.php，我希望他们留在contact.php上