3

我正在尝试将视频文件与具有一堆属性的记录相关联,但似乎不允许用户以一种形式完成所有操作 - 为视频命名,提供描述并回答一些问题,然后上传文件。

以下是我要执行的步骤:

  1. 为用户提供一个页面,该页面包含具有以下字段的表单:名称、描述、文件选择器。
  2. 该文件被存储为一个 blob,并且 id 与名称和描述一起被记录。

有没有人有任何我可以学习的例子或你可以指出我的教程?来自谷歌的那个只显示上传文件并被重定向到它。

感谢并为一个新问题感到抱歉!

4

3 回答 3

5

http://demofileuploadgae.appspot.com/ - 我到 blobstore 的演示上传器。

我的上传代码:http ://code.google.com/p/gwt-examples/source/browse/trunk/DemoUpload/src/org/gonevertical/upload/#upload/server%3Fstate%3Dclosed

于 2010-09-06T16:30:47.807 回答
3

这是我用来上传图片并将它们与文章相关联的代码。最困难的一点是让文章 id 到达上传处理程序,我通过将文件名设置为文章 id 来解决这个问题以解决这个问题。

from lib import urllib2_file
from lib.urllib2_file import UploadFile

# this view serves a task in a queue
def article(request):
       article = Article.objects.get(id=form.cleaned_data['article'])

       try:
            image = StringIO(urllib2.urlopen(image_url).read())
        except (urllib2.HTTPError, DownloadError):
            article.parsed = True
            article.save()
        else:
            image = UploadFile(image, '.'.join([str(article.id), image_url.rsplit('.', 1)[1][:4]]))
            upload_url = blobstore.create_upload_url(reverse('Articles.views.upload'))

            try:
                urllib2.urlopen(upload_url, {'file': image})
            except (DownloadError, RequestTooLargeError):
                pass

    return HttpResponse(json.dumps({'status': 'OK'}))

def upload(request):
    if request.method == 'POST':
        blobs = get_uploads(request, field_name='file', populate_post=True)

        article = Article.objects.get(id=int(blobs[0].filename.split('.')[0]))
        article.media = blobs[0].filename
        article.parsed = True
        article.save()

        return HttpResponseRedirect(reverse('Articles.views.upload'))
    else:
        return HttpResponse('meow')

    def upload(request):
        if request.method == 'POST':
            blobs = get_uploads(request, field_name='file', populate_post=True)

            article = Article.objects.get(id=int(blobs[0].filename.split('.')[0]))
            article.media = blobs[0].filename
            article.parsed = True
            article.save()

            return HttpResponseRedirect(reverse('Articles.views.upload'))
        else:
            return HttpResponse('meow')

# this serves the image
def image(request):
    blob = BlobInfo.gql("WHERE filename='%s' LIMIT 1" % request.form.cleaned_data['id'])[0]

    return HttpResponse(BlobReader(blob.key()).read(),
                        content_type=blob.content_type)

你也需要这个http://fabien.seisen.org/python/urllib2_file/

于 2010-08-04T20:19:45.427 回答
1

这是我的做法。它比你想象的更直接。请注意以下取自Blobstore 概述的内容。 “当 Blobstore 重写用户的请求时,上传文件的 MIME 部分的正文被清空,并且 blob 键作为 MIME 部分标头添加。所有其他表单字段和部分都被保留并传递给上传处理程序。” 在上传处理程序中,您可以对其他表单字段进行任何操作。

    class Topic(db.Model):
        title = db.StringProperty(multiline=False)
        blob = blobstore.BlobReferenceProperty()
        imageurl = db.LinkProperty()

    class MainHandler(webapp.RequestHandler):
        def get(self):
            upload_url = blobstore.create_upload_url('/upload')
            self.response.out.write('<html><body>')
            self.response.out.write('<form action="%s" method="POST" enctype="multipart/form-data">' % upload_url)
            self.response.out.write("""Upload File: <input type="file" name="file"><br>
            <div><label>Title:</label></div>
            <div><textarea name="title" rows="1" cols="25"></textarea></div><input type="submit" 
                name="submit" value="Submit"> </form>""")
            self.response.out.write('<br><br><h2>TOPIC LIST</h2><table border="1"><tr><td>')
            for topic in Topic.all():                
                self.response.out.write('<div><img src="%s=s48"/>' % topic.imageurl)
                self.response.out.write('<div><b>Image URL: </b><i>%s</i></div>' % topic.imageurl)
                self.response.out.write('<div><b>Title: </b><i>%s</i></div>' % topic.title)
            self.response.out.write('</td></tr></table><br>') 
            self.response.out.write('</body></html>')

    class UploadHandler(blobstore_handlers.BlobstoreUploadHandler):
        def post(self):
            upload_files = self.get_uploads('file')  # 'file' is file upload field in the form
            blob_info = upload_files[0]
            topic = Topic()
            topic.title = self.request.get("title")
            topic.blob = blob_info.key()
            topic.imageurl = images.get_serving_url(str(blob_info.key()))
            topic.put()        
            self.redirect('/')
def main():
    application = webapp.WSGIApplication(
          [('/', MainHandler),
           ('/upload', UploadHandler),
          ], debug=True)
    run_wsgi_app(application)
于 2010-11-12T07:29:27.940 回答