有使用Uploadifive和Django的经验吗？

Question

我正试图在Django实施Uploadifive。这对我来说很困难，因为我和Django的关系不是很好。一点都没有。当然，这应该告诉我，我应该用这个等着，先学到更多的Django，但是..。是的，我无法远离它..。

据我所知，我需要看看Uploadifive包附带的PHP脚本，并在Django视图中编写相应的脚本。这是我的猜测至少..。问题是，我似乎在网上找不到任何指南，也找不到任何关于如何做的提示。有谁有这样的例子，我可以看看吗？或者我该去哪有什么建议吗？

到目前为止，我已经在urls.py中创建了一种模式，该模式将浏览器定向到将用户发送到正确网页的views.py def。JavaScript在站点上初始化(我可以看到“选择文件”按钮)，但是当我选择图像时，什么都不会发生。我试图根据我找到的基于flash的Uploadify指南构建一个视图脚本，但这是行不通的。

编辑:有人指出，很自然地有必要看看PHP代码实际上做了什么。它确实是一个付费软件，但我无法想象PHP代码在代码中占这么大的一部分，以至于它会“泄露”任何东西。真正的工作在于javascript文件。

PHP上传收件人脚本：

<?php
/*
UploadiFive
Copyright (c) 2012 Reactive Apps, Ronnie Garcia
*/

// Set the uplaod directory
$uploadDir = '/uploads/';

// Set the allowed file extensions
$fileTypes = array('jpg', 'jpeg', 'gif', 'png'); // Allowed file extensions

if (!empty($_FILES)) {
    $tempFile   = $_FILES['Filedata']['tmp_name'];
    $uploadDir  = $_SERVER['DOCUMENT_ROOT'] . $uploadDir;
    $targetFile = $uploadDir . $_FILES['Filedata']['name'];

    // Validate the filetype
    $fileParts = pathinfo($_FILES['Filedata']['name']);
    if (in_array(strtolower($fileParts['extension']), $fileTypes)) {

        // Save the file
        move_uploaded_file($tempFile,$targetFile);
        echo 1;
    } else {
        // The file type wasn't allowed
        echo 'Invalid file type.';
    }
}
?>

模板：

    {% extends 'base/index.html' %}
    {% block stylesheet %}
      <link rel="stylesheet" type="text/css" href="{{ STATIC_URL }}uploadifive/uploadifive.css">
      <script type="text/javascript" src="http://code.jquery.com/jquery-1.7.2.min.js"></script>
      <script type="text/javascript" src="{{ STATIC_URL }}uploadifive/jquery.uploadifive.js"></script>
      <script type="text/javascript">
      $(document).ready(function(){
        $('#file_upload').uploadifive({
          'debug': true,
          'formData':{'test':'something'},
          'uploadScript': '/upload/',
          'queueID': 'queue',
          'cancelImage': '{{ STATIC_URL }}uploadifive/uploadifive-cancel.png',
          'auto': true
        });
      });
      </script>
    {% endblock %}

    {% block content %}
      <form>
        <input type="file" name="file_upload" id="file_upload" />
      </form>
    {% endblock %}

views.py

    from django.shortcuts import render
    from gallery.models import Pic,Comment,Tag
    from django.conf import settings
    from django.http import HttpResponse
    from django.shortcuts import render

    def upload(request):
        if request.method == 'POST':
            for field_name in request.FILES:
                uploaded_file = request.FILES[field_name]

                #write the file into destination
                destination_path = '<absolute path to project>/media/gallery/pictures/%s' % (uploaded_file.name)
                destination = open(destination_path, 'wb+')
                for chunk in uploaded_file.chunks():
                    destination.write(chunk)
                destination.close()

            #indicate that everything is OK
            return HttpResponse("ok", mimetype="text/plain")
        else:
            #show the upload UI
            return render(request, 'gallery/upload.html')

Answer 1

下面是PHP文件的Python翻译：

# Set the uplaod directory
upload_dir = '/absolute/path/to/upload/dir/'

# Set the allowed file extensions
file_types = ['jpg', 'jpeg', 'gif', 'png'] # Allowed file extensions

if len(request.FILES):
    target_file = upload_dir + request.FILES['field_name'].name

    # Validate the filetype
    filename, ext = os.path.splitext(request.FILES['field_name'].name)
    if ext.lower() in file_types:

        # Save the file
        with open(target_file, 'wb+') as destination:
            for chunk in request.FILES['field_name'].chunks():
                destination.write(chunk)
    else:
        # The file type wasn't allowed
        print 'Invalid file type.'

如果一次发送多个文件，最好将代码包装在一个for循环中：

for file in request.FILES.values():
    target_file = upload_dir + file.name
    ...

然后，您显然希望返回正确的响应对象并更好地处理错误，但是您应该能够自己处理。