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how to upload muti file using html5 on google app engine(python)

开发者 https://www.devze.com 2023-01-26 14:37 出处:网络
this is my code, the upload.py: class MyModel(db.Model): data = db.BlobProperty(required=True) mimetype = db.StringProperty(required=True)

this is my code, the upload.py:

class MyModel(db.Model):
    data = db.BlobProperty(required=True)
    mimetype = db.StringProperty(required=True)

class BaseRequestHandler(webapp.RequestHandler):
  def render_template(self, filename, template_args=None):
    if not template_args:
      template_args = {}
    path = os.path.join(os.path.dirname(__file__), 'templates', filename)
    self.response.out.write(template.render(path, template_args))

class upload(BaseRequestHandler):
  def get(self):
    self.render_template('index.html',)
  def post(self):
    file = self.request.POST['file']
    obj = MyModel(data=file.value, mimetype=file.type)
    obj.put()
    o=file
    #self.response.out.write(''.join('%s: %s <br/>' % (a, getattr(o, a)) for a in dir(o)))
    #return
    file_url = "http://%s/download/%d/%s" % (self.request.host, obj.key().id(), file.filename)
    self.response.out.write("Your uploaded file is now available at <a href='%s'>%s</a>" % (file_url,file_url,))

and the index.html is :

<form enctype="multipart/form-data" action="/" method="post">
  <input type="file" name="file" multiple="true" />
  <input type="submit" />
</form>

you can use file = self.request.POST['file'] to get 开发者_如何转开发one file , but using html5 muti file ,

how to get muti file using python when post ?

thanks

upload

it is ok now , Follow this article: Receive multi file post with google app engine

class Download_file(db.Model):
    data = db.BlobProperty(required=True)
    mimetype = db.StringProperty(required=True)
    download_url = db.StringProperty()

class BaseRequestHandler(webapp.RequestHandler):
  def render_template(self, filename, template_args=None):
    if not template_args:
      template_args = {}
    path = os.path.join(os.path.dirname(__file__), 'templates', filename)
    self.response.out.write(template.render(path, template_args))

class upload(BaseRequestHandler):
    def get(self):
        files=Download_file.all()
        self.render_template('index.html',{'files':files})
    def post(self):
        files = self.request.POST.multi.__dict__['_items']
        #self.response.out.write(files)
        for file in files:
            file=file[1]
            obj = Download_file(data=file.value, mimetype=file.type)
            obj.put()
            file_url = "http://%s/download/%d/%s" % (self.request.host, obj.key().id(), file.filename)
            file_url = "<a href='%s'>%s</a>" % (file_url,file_url,)
            obj.download_url=file_url
            obj.put()
            self.response.out.write("Your uploaded file is now available at %s </br>" % (file_url))

class download(BaseRequestHandler):
    def get(self,id,filename):
        #id=self.request.get('id')
        entity = Download_file.get_by_id(int(id))
        self.response.headers['Content-Type'] = entity.mimetype
        self.response.out.write(entity.data)


You probably want to use the blobstore service for this; I wrote a series of posts (1, 2, 3) covering how to do multiple-file upload to the blobstore, using an upload widget.


AFAIK, the request.POST['file'] should be a dictionary of files, i.e. POST['file'] should have keys as names of the files uploaded and the value should be the contents of the files respectively, i.e. POST['file']['avatar.png'] = ... # raw image data.

I don't know what's the functionality provided by the GAE HTTP Request class, but it should be consistent to this. Whatever the case, it's definitely in self.request, somwhere!

EDIT:

Ok, I just noticed GAE creates a file object for you, my guessing is this should work:

for file in POST['file']:
    obj = MyModel(data=file.value, mimetype=file.type)
    obj.put()
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