PHP AJAX接触形式不发送下拉框中选择电子邮件

Question

试图编辑联系人窗体上的HTML主题，已经成功迄今除了“mydropdown”投寄箱时输入数据，并提交了不正确填充。我在这里做错了什么？

这里是我的home.js文件中的Ajax功能：

// ------------------------------------------------------------- 
// Contact form 
// ------------------------------------------------------------- 

//Ajax contact 

    $(function() { 
    $('#contact-form').submit(function() { 
    var name = $('#name').val(); 
    var email =$('#email').val(); 
    var subject =$('#subject').val(); 
    var message =$('#message').val(); 
    var dates =$('#dates').val(); 
    var mydropdown =$('#mydropdown').val(); 
    var phone =$('#phone').val(); 
    var pet =$('#pet').val(); 
      $.ajax({ 
       type: 'POST', 
       url: './sendemail.php', 
       data: { name:name, email:email, subject:subject, pet:pet, 
dates:dates, mydropdown:mydropdown, phone:phone, message:message } 
      }); 
      $('#contact-form').trigger('reset'); 
      $('.thank-you').addClass("show-on-success"); 
      return false; 
     }); 
    }); 


}); // end custom jQuery 

，这里是我的PHP文件 “sendemail.php”

 <?php 
     $name = $_POST['name']; 
     $email = $_POST['email']; 
     $message = $_POST['message']; 
     $phone = $_POST['phone']; 
     $pet = $_POST['pet']; 
     $dates = $_POST['dates']; 
     $mydropdown = $_POST['mydropdown'] 
    $formcontent="From: $name \n Phone: $phone \n Pet's Name: $pet \n 
     Drop Off Dates: $dates \n Location: $mydropdown \n 
       Services: $message"; 
        $recipient = "alpham4le@gmail.com"; 
       $subject = "Contact Form"; 
       $mailheader = "From: $email \r\n"; 
    mail($recipient, $subject, $formcontent, $mailheader) or die("Error!"); 

     ?> 

这里是我的索引HTML我的实际形式文件：

<!-- contact-form --> 

       </div> 
      </div><!-- /row --> 

      <!-- row --> 
      <div class="row"> 
       <div class="col-xs-12"> 
        <!-- contact-form --> 
        <form id="contact-form" class="contact-form" 
name="contact-form" method="post" action="#"> 
         <div class="row"> 
          <div class="form-group col-sm-6"> 
           <input type="text" id="name" name="name" 
required class="form-control wow fadeInUp" placeholder="Your Name"> 
           <input type="text" id="phone" name="phone" 
required class="form-control wow fadeInUp" placeholder="Your Phone Number"> 
           <input type="text" id="email" name="email" 
required class="form-control wow fadeInUp" placeholder="Your Email Address"> 
           <input type="text" id="pet" name="pet" 
required class="form-control wow fadeInUp" placeholder="Your Pet's Name"> 
           <input type="text" id="dates" name="dates" 
required class="form-control wow fadeInUp" placeholder="Drop Off Dates"> 

          </div> 
          <select name="mydropdown"> 
           <option value="Location">Location</option> 
           <option value="Downtown">Downtown</option> 
           <option value="Valley">Valley</option> 
                 </select> 
                 <br /> 
          <div class="form-group col-sm-6"> 
           <textarea name="message" id="message" 
required class="form-control wow fadeInUp" rows="30" placeholder="Services 
you're interested in & any other questions.."></textarea> 
          </div> 

         <div class="form-group col-sm-12"> 
          <h1 class="thank-you"><span>Thank you! Your mail 
has been sent!</span></h1>     
          <button type="submit" class="btn btn-angle wow 
zoomIn">Submit Message</button> 
         </div> 

         </div>             

        </form> <!-- contact-form --> 

我在做什么错在这里？我认为它与ajax表单有关，其他所有工作都很好，只是位置：表单的一部分没有发送任何下拉选项。

Answer 1

你没有id="mydropdown"

<select name="mydropdown"> 
    <option value="Location">Location</option> 
    <option value="Downtown">Downtown</option> 
    <option value="Valley">Valley</option> 
</select> 

所以应该

<select name="mydropdown" id="mydropdown"> 
    <option value="Location">Location</option> 
    <option value="Downtown">Downtown</option> 
    <option value="Valley">Valley</option> 
</select> 

Answer 2

你的选择没有一个ID为它设置

<select name="mydropdown"> 

修改它是

<select name="mydropdown" id="mydropdown"> 

检查您的控制台会帮你出这一点。

在jQuery的你有一个变种 var mydropdown =$('#mydropdown').val();其与下降的ID下来，在你的HTML结合