Question

我已经查看了很多问题，但我似乎无法找到解决我具体问题的问题。

我正在尝试允许用户（在我的Google App Engine网站上）将视频上传到wistia（视频主机）。

我可以通过直接打开文件来使其工作，但这仅在文件已经在app文件夹中时才有效：

class VideoUploadHandler(webapp2.RequestHandler):
    def post(self):
        title = self.request.get('title')
        video = self.request.get('video')
        description = self.request.get('description')

        url = 'https://upload.wistia.com'
        params = {
            'api_password'    : 'my_password',
            'file'            : open(video, "rb"),  #<--- HOW TO USE FILE OBJECT?
            'project_id'      : 'my_project_id',
            'name'            : title,
            'description'     : description,
            }
        opener = urllib2.build_opener(MultipartPostHandler.MultipartPostHandler) 
        response = opener.open(url, params)

        logging.warning(response.read())

        self.redirect("https://www.my_webpage.com/")

如果变量'video'仅仅是文件名并且文件位于同一文件夹中，则此方法有效，但我显然需要允许用户选择任何文件，因此我将HTML更改为：

HTML：

<form enctype="multipart/form-data" action="/video_upload/" method="post">
<p>Title: <input type="text" name="title"/></p>
<p><input type="file" name="video"/></p>
<p>Description:<br><textarea rows="4" cols="50" name="description"></textarea></p>
<p><input type="submit" value="Upload"/></p>
</form>

但是，我无法理解如何在POST请求中使用文件对象。出于某种原因，我无法从已发布的其他答案中收集这些信息。

如何使用此HTML表单方法构建我的'params'？

修改

为了澄清，我现在有了这个简单的HTML：

<input id="video_input" type="file" name="video"/>
<a id="video_button">SUBMIT VIDEO</a>

此JavaScript基于this和this：

$('#video_button').click(function() {
    var file = new FormData();
    file.append('video', $('#video_input').get(0).files[0]);
    $.ajax({
        type: "POST",
        url: "/video_upload/",
        data: file,
        processData: false,
        contentType: false
    });
});

但我仍然没有找到任何充分解释如何获取我现在在Python中的文件对象并将其正确传递给wistia POST参数的任何内容。任何想法或链接？

修改使用时回溯：

params = {
            'api_password'    : 'my_password',
            'file'            : video,
            'project_id'      : 'my_project_id',
            'name'            : title,
            'description'     : description,
            }

（video.py）是带有VideoUploadHandler类的文件。

WARNING  2015-09-05 07:13:16,970 urlfetch_stub.py:504] Stripped prohibited headers from URLFetch request: ['Content-Length', 'Host']
ERROR    2015-09-05 12:13:21,026 webapp2.py:1552] HTTP Error 500: Internal Server Error

Traceback (most recent call last):

  File "C:\...\webapp2-2.5.2\webapp2.py", line 1535, in __call__

    rv = self.handle_exception(request, response, e)

  File "C:\...\webapp2-2.5.2\webapp2.py", line 1529, in __call__

    rv = self.router.dispatch(request, response)

  File "C:\...\webapp2-2.5.2\webapp2.py", line 1278, in default_dispatcher

    return route.handler_adapter(request, response)

  File "C:\...\webapp2-2.5.2\webapp2.py", line 1102, in __call__

    return handler.dispatch()

  File "C:\...\webapp2-2.5.2\webapp2.py", line 572, in dispatch

    return self.handle_exception(e, self.app.debug)

  File "C:\...\webapp2-2.5.2\webapp2.py", line 570, in dispatch

    return method(*args, **kwargs)

  File "C:\...\video.py", line 84, in post

    response = opener.open(url, params)

  File "C:\...\urllib2.py", line 410, in open

    response = meth(req, response)

  File "C:\...\urllib2.py", line 523, in http_response

    'http', request, response, code, msg, hdrs)

  File "C:\...\urllib2.py", line 448, in error

    return self._call_chain(*args)

  File "C:\...\urllib2.py", line 382, in _call_chain

    result = func(*args)

  File "C:\...\urllib2.py", line 531, in http_error_default

    raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)

HTTPError: HTTP Error 500: Internal Server Error

INFO     2015-09-05 07:13:21,700 module.py:788] default: "POST /video_upload/ HTTP/1.1" 500 2253

Answer 1

编辑：试试这个：

from poster.encode import multipart_encode, MultipartParam
from google.appengine.api import urlfetch

class VideoUploadHandler(webapp2.RequestHandler):
    def post(self):
        title = self.request.get('title')
        description = self.request.get('description')
        payload = {
            'api_password'    : 'my_password',
            'project_id'      : 'my_project_id',
            'name'            : title,
            'description'     : description,
            }
        file_data = self.request.POST['video']
        payload['file'] = MultipartParam('file', filename=file_data.filename,
                                              filetype=file_data.type,
                                              fileobj=file_data.file)
        data,headers= multipart_encode(payload)
        send_url = "https://upload.wistia.com"
        resp = urlfetch.fetch(url=send_url, payload="".join(data), method=urlfetch.POST, headers=headers)       
        logging.info(resp)
        self.redirect("/")

您需要这个lib，这是根据此post

改编的

HTML，jQuery，Python - 表单文件上传（GAE）

1 个答案: