PHP无法获取从AJAX传递的POST值

我按照这篇文章提到的内容上传文件。有一点不同的是，我在表单上又添加了一个文本输入字段。图像文件成功上传到服务器，但输入字段中的值似乎没有传递给PHP进行数据库更新。数据库函数被触发，数据库记录被添加，但缺少表单中的值。

有人能指出我遗漏了什么吗？谢谢

$.validate({
  form: '#frmSlide',
  modules: 'file, html5',
  validateOnBlur: false,
  errorMessagePosition: 'top', // Instead of 'element' which is default
  scrollToTopOnError: false, // Set this property to true if you have a long form
  onSuccess: function($form) {
    var file_data = $('#a_imgfile').prop('files')[0];
    var form_data = new FormData();
    form_data.append('file', file_data);
    $.ajax({
      url: 'slide_upd.php', // point to server-side PHP script
      dataType: 'text', // what to expect back from the PHP script, if anything
      cache: false,
      contentType: false,
      processData: false,
      data: form_data,
      type: 'post',
      success: function(data) {
        alert(data);
      }
    });
  }
});

<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post" enctype="multipart/form-data" class="form-horizontal" id="frmSlide">
  <div class="form-group">
    <label class="col-sm-4 control-label" for="imgfile">Image file</label>
    <div class="col-sm-8">
      <input type="file" id="a_imgfile" data-validation="required mime size" data-validation-allowing="jpg, png, gif" data-validation-ratio="1:1" data-validation-max-size="1M" data-validation-error-msg="An image file is mandatory." />
    </div>
  </div>
  <div class="form-group">
    <div class="col-sm-8 col-md-offset-4" id="image-holder">
    </div>
  </div>
  <div class="form-group">
    <label class="col-sm-4 control-label" for="seq">Sequence</label>
    <div class="col-sm-8">
      <input class="form-control server" name="a_seq" id="a_seq" type="number" min="1" max="4" value="" placeholder="Enter display sequence of this slide" data-validation-error-msg="Only 1 to 4 is allowed." />
    </div>
  </div>
  <div class="form-group">
    <div class="col-sm-offset-4 col-sm-8">
      <button name="btnUpd" id="btnUpd" type="submit" class="clsUpd btn btn-primary"><i class="fa fa-floppy-o"></i>&nbsp;Update</button>
    </div>
  </div>
</form>

<?php
	$image_name = $_FILES['file']['name'];
	$image_size = $_FILES['file']['size'];
	$image_temp = $_FILES['file']['tmp_name'];
	
	move_uploaded_file($image_temp, 'img/'.$image_name);
	$seq = $_POST['a_seq'];
	addSlide($seq);
?>

function addSlide($seq) {
	$seq = (int)$seq;
	mysql_query("INSERT INTO slide (seq, lastchgat) 
		VALUES ('$seq', now())")  or die(mysql_error());
}