Question

我有一个简单的部分，用户可以在其中上传文件，现在我想在文件成功上传后显示成功消息

HTML

 <form id="uploadform" action="upload.php" method="post" enctype="multipart/form-data">
              <label for="file"><span>Filename:</span></label>
              <input type="file" id="fileupload"  name="file" id="file" /> 
              <br />
              <input id="#submit_cf" type="submit" name="submit" class="btn btn-primary" value="Upload file" />
            </form>
            <span id="success_message"></span>

Ajax;

 function doSuccess(acton,message,disable)
        {
            $(acton).show();
            $(acton+' h2').text(message);
            $(disable).attr('disabled','disabled');
        }

    $('#uploadform').on('submit',function(e){
        // This is fine, it does prevent the form from submitting
        e.preventDefault();

        var form_data = $('#fileupload').prop('files')[0]; 
            console.log(form_data);
        // Run messaging
        doSuccess('#success_message','Processing, please wait...','#submit_cf');
        // Run ajax
        $.ajax({
            url : "upload.php",
            type: "POST",
            data: form_data,
            success: function(response) {
                $('#success_message ').text(response);
            }
        });
    });

在控制台console.log(form_data);上

返回此

这里是upload.php

<?php 

$allowedExts = array("jpg", "jpeg", "gif", "png", "mp3", "mp4", "wma");

$extension = pathinfo($_FILES['file']['name'], PATHINFO_EXTENSION);

if ((($_FILES["file"]["type"] == "video/mp4")
|| ($_FILES["file"]["type"] == "audio/mp3")
|| ($_FILES["file"]["type"] == "audio/wma")
|| ($_FILES["file"]["type"] == "image/pjpeg")
|| ($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg"))

&& ($_FILES["file"]["size"] < 5000000)
&& in_array($extension, $allowedExts))

  {
  if ($_FILES["file"]["error"] > 0)
    {
    echo "Return Code: " . $_FILES["file"]["error"] . "<br />";
    }
  else
    {
    echo "Upload: " . $_FILES["file"]["name"] . "<br />";
    echo "Type: " . $_FILES["file"]["type"] . "<br />";
    echo "Size: " . ($_FILES["file"]["size"] / 1024) . " Kb<br />";
    echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br />";

    if (file_exists("upload/" . $_FILES["file"]["name"]))
      {
      echo $_FILES["file"]["name"] . " already exists. ";

      }
    else
      {
      move_uploaded_file($_FILES["file"]["tmp_name"],
      "upload/" . $_FILES["file"]["name"]);
      echo "Stored in: " . "upload/" . $_FILES["file"]["name"];
      }
    }
  }
else
  {
  echo "Invalid file";
  }
?>

现在，当我上传文件时，出现以下错误

jquery.min.js:4 Uncaught TypeError: Illegal invocation

我的代码有什么问题？

Answer 1

您正在将文件传递到return this.http.post(this.env.STORES_URL, { name: name, description: description, phone: phone, province_id: province_id, kota_id: kota_id, address: address, logo: logo, banner: banner });，并且jQuery在尝试将其编码为表单数据时出错。

您应该：

改为传递真实形式的数据对象
告诉jQuery不处理数据
告诉jQuery不设置Content-Type（浏览器将使用FormData对象自动进行设置）

因此：

data

PHP和Ajax：非法调用

1 个答案: