PHP无法获取从AJAX传递的POST值

Question

我正在关注这篇文章提到的上传文件。 略有不同是我在表单上添加了另一个文本输入字段。 图像文件已成功上传到服务器，但是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，并且数据库记录已添加，但缺少表单中的值。

谁能指出我错过的事情？ 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()); } 

Answer 1

1. a_seq不会附加到form_data。
2. 添加var a_seq = $('#a_seq').val();
3. form_data.append（'a_seq'，a_seq）;
4. 应该很好走

Answer 2

我认为这可以解决您的问题

$.validate({
  form: '#frmSlide',
  modules: 'file, html5',
  validateOnBlur: false,
  errorMessagePosition: 'top', // Instead of 'element' which is default
  scrollToTopOnError: false, // Set this property to true if you have a long form
  onSuccess: function($form) {
    //var file_data = $('#a_imgfile').prop('files')[0];
    //var form_data = new FormData();
    //form_data.append('file', file_data);
    //------ instead of three lines i just did this and works fine for me -------
    var formData=new FormData($('#frmSlide')[0]);
    $.ajax({
      url: 'slide_upd.php', // point to server-side PHP script
      cache: false,
      contentType: false,
      processData: false,
      data: formData,
      type: 'post',
      success: function(data) {
        alert(data);
      }
    });
  }
});