So I am trying to create a binary file and save it into my database. I am using REDIS and SQLALCHEMY as a framework for my database. I can use send_file to send the actual file whenever the user accesses a URL but how do I make sure that the file is saved in the route and it could stay there every time a user accesses the URL.
I am sending the file from a client-python it's not in my
directory
what I need in a nutshell is to save the file from the client-python to a database to "downloadable" it to the browser-client so it would actually be available for the browser-client is there any way of doing this? Maybe a different way that I didn't think about
I had to encode the data with base64, send it to the database and then decode it and send the file as binary data.
I'm using twitter from python in an environment where I can't store files.
I get a HTTP POST with a text and an image and want to create a tweet from this data without writing a local file (it's zappa on AWS api environment).
Tweepy only allows filenames, which does not work for me.
python-twitter seems to have something like that, but I can't find a doc for this.
Should I just send POST requests to twitter for uploading the images? Is there a simpler way?
Try passing a io.BytesIO to tweepy's API.update_with_media as file.
filename – The filename of the image to upload. This will automatically be opened unless file is specified
...
file – A file object, which will be used instead of opening filename. filename is still required, for MIME type detection and to use as a form field in the POST data
Edit:
It looks like you have the image data base64 encoded. You can use base64.b64decode to decode it before creating the io.BytesIO:
file = io.BytesIO(base64.b64decode(base64_data))
I've noticed that the FTP library doesn't seem to have a method or function of straight up downloading a file from an FTP server. The only function I've come across for downloading a file is ftp.retrbinary and in order to transfer the file contents, you essentially have to write the contents to a pre-existing file on the local computer where the Python script is located.
Is there a way to download the file as-is without having to create a local file first?
Edit: I think the better question to ask is: do I need to have a pre-existing file in order to download an FTP server file's contents?
To download a file from FTP this code will do the job
import urllib urllib.urlretrieve('ftp://server/path/to/file', 'file') # if you need to pass credentials: # urllib.urlretrieve('ftp://username:password#server/path/to/file', 'file')
I'm looking for a way to sell someone a card at an event that will have a unique code that they will be able to use later in order to download a file (mp3, pdf, etc.) only one time and mask the true file location so a savvy person downloading the file won't be able to download the file more than once. It would be nice to host the file on Amazon S3 to save on bandwidth where our server is co-located.
My thought for the codes would be to pre-generate the unique codes that will get printed on the cards and store those in a database that could also have a field that stores the number of times the file was downloaded. This way we could set how many attempts we would allow the user for downloading the file.
The part that I need direction on is how do I hide/mask the original file location so people can't steal that url and then download the file as many times as they want. I've done Google searches and I'm either not searching using the right keywords or there aren't very many libraries or snippets out there already for this type of thing.
I'm guessing that I might be able to rig something up using django.views.static.serve that acts as a sort of proxy between the actual file and the user downloading the file. The only drawback to this method I would think is that I would need to use the actual web server and wouldn't be able to store the file on Amazon S3.
Any suggestions or thoughts are greatly appreciated.
Neat idea. However, I would warn against the single-download method, because there is no guarantee that their first download attempt will be successful. Perhaps use a time-expiration method instead?
But it is certainly possible to do this with Django. Here is an outline of the basic approach:
Set up a django url for serving these files
Use a GET parameter which is a unique string to identify which file to get.
Keep a database table which has a FileField for the file to download. This table maps the unique strings to the location of the file on the file system.
To serve the file as a download, set the response headers in the view like this:
(path is the location of the file to serve)
with open(path, 'rb') as f:
response = HttpResponse(f.read())
response['Content-Type'] = 'application/octet-stream';
response['Content-Disposition'] = 'attachment; filename="%s"' % 'insert_filename_here'
return response
Since we are using this Django page to serve the file, the user cannot find out the original file location.
You can just use something simple such as mod_xsendfile. This functionality is also available in other popular webservers such lighttpd or nginx.
It works like this: when enabled your application (e.g. a trivial PHP script) can send a special response header, causing the webserver to serve a static file.
If you want it to work with S3 you will need to handle each and every request this way, meaning the traffic will go through your site, from there to AWS, back to your site and back to the client. Does S3 support symbolic links / aliases? If so you might just redirect a valid user to one of the symbolic URLs and delete that symlink after a couple of hours.
I am using this file storage engine to store files to Amazon S3 when they are uploaded:
http://code.welldev.org/django-storages/wiki/Home
It takes quite a long time to upload because the file must first be uploaded from client to web server, and then web server to Amazon S3 before a response is returned to the client.
I would like to make the process of sending the file to S3 asynchronous, so the response can be returned to the user much faster. What is the best way to do this with the file storage engine?
Thanks for your advice!
I've taken another approach to this problem.
My models have 2 file fields, one uses the standard file storage backend and the other one uses the s3 file storage backend. When the user uploads a file it get's stored localy.
I have a management command in my application that uploads all the localy stored files to s3 and updates the models.
So when a request comes for the file I check to see if the model object uses the s3 storage field, if so I send a redirect to the correct url on s3, if not I send a redirect so that nginx can serve the file from disk.
This management command can ofcourse be triggered by any event a cronjob or whatever.
It's possible to have your users upload files directly to S3 from their browser using a special form (with an encrypted policy document in a hidden field). They will be redirected back to your application once the upload completes.
More information here: http://developer.amazonwebservices.com/connect/entry.jspa?externalID=1434
There is an app for that :-)
https://github.com/jezdez/django-queued-storage
It does exactly what you need - and much more, because you can set any "local" storage and any "remote" storage. This app will store your file in fast "local" storage (for example MogileFS storage) and then using Celery (django-celery), will attempt asynchronous uploading to the "remote" storage.
Few remarks:
The tricky thing is - you can setup it to copy&upload, or to upload&delete strategy, that will delete local file once it is uploaded.
Second tricky thing - it will serve file from "local" storage until it is not uploaded.
It also can be configured to make number of retries on uploads failures.
Installation & usage is also very simple and straightforward:
pip install django-queued-storage
append to INSTALLED_APPS:
INSTALLED_APPS += ('queued_storage',)
in models.py:
from queued_storage.backends import QueuedStorage
queued_s3storage = QueuedStorage(
'django.core.files.storage.FileSystemStorage',
'storages.backends.s3boto.S3BotoStorage', task='queued_storage.tasks.TransferAndDelete')
class MyModel(models.Model):
my_file = models.FileField(upload_to='files', storage=queued_s3storage)
You could decouple the process:
the user selects file to upload and sends it to your server. After this he sees a page "Thank you for uploading foofile.txt, it is now stored in our storage backend"
When the users has uploaded the file it is stored temporary directory on your server and, if needed, some metadata is stored in your database.
A background process on your server then uploads the file to S3. This would only possible if you have full access to your server so you can create some kind of "deamon" to to this (or simply use a cronjob).*
The page that is displayed polls asynchronously and displays some kind of progress bar to the user (or s simple "please wait" Message. This would only be needed if the user should be able to "use" (put it in a message, or something like that) it directly after uploading.
[*: In case you have only a shared hosting you could possibly build some solution which uses an hidden Iframe in the users browser to start a script which then uploads the file to S3]
You can directly upload media to the s3 server without using your web application server.
See the following references:
Amazon API Reference : http://docs.amazonwebservices.com/AmazonS3/latest/dev/index.html?UsingHTTPPOST.html
A django implementation : https://github.com/sbc/django-uploadify-s3
As some of the answers here suggest uploading directly to S3, here's a Django S3 Mixin using plupload:
https://github.com/burgalon/plupload-s3mixin
I encountered the same issue with uploaded images. You cannot pass along files to a Celery worker because Celery needs to be able to pickle the arguments to a task. My solution was to deconstruct the image data into a string and get all other info from the file, passing this data and info to the task, where I reconstructed the image. After that you can save it, which will send it to your storage backend (such as S3). If you want to associate the image with a model, just pass along the id of the instance to the task and retrieve it there, bind the image to the instance and save the instance.
When a file has been uploaded via a form, it is available in your view as a UploadedFile file-like object. You can get it directly out of request.FILES, or better first bind it to your form, run is_valid and retrieve the file-like object from form.cleaned_data. At that point at least you know it is the kind of file you want it to be. After that you can get the data using read(), and get the other info using other methods/attributes. See https://docs.djangoproject.com/en/1.4/topics/http/file-uploads/
I actually ended up writing and distributing a little package to save an image asyncly. Have a look at https://github.com/gterzian/django_async Right it's just for images and you could fork it and add functionalities for your situation. I'm using it with https://github.com/duointeractive/django-athumb and S3