发送空值的邮件函数

Question

我正在尝试使用phpmail()函数向用户帐户发送电子邮件。邮件发送成功了，但问题是它向我发送的是没有值的空白电子邮件！发送电子邮件的联系人页面的代码如下：-

<form  class="contact-form"  method="POST" action="sendemail.php">
                  <div class="row-fluid">
                    <div class="span5">
                        <label>First Name</label>
                        <input type="text" class="input-block-level" required="required" placeholder="Your First Name" name="name">
                        <label>Last Name</label>
                        <input type="text" class="input-block-level" required="required" placeholder="Your Last Name" name="lname">
                        <label>Email Address</label>
                        <input type="text" class="input-block-level" required="required" placeholder="Your email address" name="email">
                    </div>
                    <div class="span7">
                        <label>Message</label>
                        <textarea name="message" id="message" required="required" class="input-block-level" rows="8" name="message"></textarea>
                    </div>

                </div>
                <input type="submit" class="btn btn-primary btn-large pull-right" value="Send Message" />
                <p> </p>

            </form>

和sendemail.php页面代码如下：

<?php
    header('Content-type: application/json');
    $status = array(
        'type'=>'success',
        'message'=>'Email sent!'
    );

    $name = @trim(stripslashes($_POST['name'])); 
    $email = @trim(stripslashes($_POST['email'])); 
    $subject = "An enquiry sir"; 
    $message = @trim(stripslashes($_POST['message'])); 

    $email_from = $email;
    $email_to = '[e-mail]';

    echo $body = 'Name: ' . $name . "\n\n" . 'Email: ' . $email . "\n\n" . 'Subject: ' . $subject . "\n\n" . 'Message: ' . $message;

    $success = @mail($email_to, $subject, $body, 'From: <'.$email_from.'>');

    echo json_encode($status);
    die;
   ?>  

为什么我在电子邮件id中得到的输出为空，例如：

姓名：

表单是使用以下JavaScript通过AJAX提交的：

var form = $('.contact-form');
form.submit(function () {
  $this = $(this);
  $.post($(this).attr('action'), function(data) {
    $this.prev().text(data.message).fadeIn().delay(3000).fade‌​Out(); 
  },'json');
  return false; 
});

Answer 1

表单提交代码未提交表单数据。下面是您提供的代码：

var form = $('.contact-form');
form.submit(function () {
  $this = $(this);
  $.post($(this).attr('action'), function(data) {
    $this.prev().text(data.message).fadeIn().delay(3000).fade‌​Out(); 
  },'json');
  return false; 
});

这就是它应该是的：

var form = $('.contact-form');
form.submit(function () {
  $this = $(this);
  $.post($(this).attr('action'), $(this).serialize(), function(data) {
    $this.prev().text(data.message).fadeIn().delay(3000).fade‌​Out(); 
  },'json');
  return false; 
});