python requests return file-like object for streaming

Question

I have created a requests object like this:

 obj.mp3 = requests.get('http://foo.com/bar.mp3', stream=True)

I thought that I could just feed this obj.mp3 object into any audio player that expects a file or an URI, obviously this idea is wrong: nothing played. Below are the full code:

#views.py

class ArticleDetailView(DetailView):
    model = Article
    template_name = 'song.html'

    def get_object(self):
        obj = super(ArticleDetailView, self).get_object()

        #code to store mp3 url in obj.mp3 and other stuff

        if obj.mp3:
            obj.mp3 = requests.get(obj.mp3, stream=True).content
        return obj

#song.html
<div class="audio">
  <audio src={{ article.mp3 }} type="audio/mpeg"> 
</div>

What is the correct way of treating return from requests as something that I can stream with a player? I know at least I can write the obj.mp3 to a file, then just point the player to the file location, but I am trying to avoid write the file to disk.

Thanks,

Innuendo · Accepted Answer · 2020-03-25 09:58:32Z

29

There's an attribute Response.raw, which is already a file-like object.

resp = requests.get(url, stream=True)
resp.raw # is what you need

Using io.BytesIO(resp.content) is not preferable since behind the scenes you're reading the same amount of data twice (also memory-wise): accessing resp.content reads everything from the network stream, then io.BytesIO(resp.content) is allocating again the same amount of memory, and then you read it from BytesIO object.

answered Mar 25, 2020 at 9:58

Innuendo

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4 Comments

JojOatXGME Over a year ago

Note that Response.raw will not respect transfer encodings as documented here. Therefore, using raw might be wrong depending on the web server.

JojOatXGME Over a year ago

After reading issue #465, I'm not sure if the problem mentioned in my last comment still exists.

JojOatXGME Over a year ago

Looks like you can add resp.raw.decode_content = True to fix it.

alanjds Over a year ago

Setting raw.decode_content had not worked for me. Wrapping with io.TextIOWrapper(resp.raw, 'utf-8') worked.

dnll · Accepted Answer · 2014-02-12 01:01:14Z

10

Look into the io module for using file-like objects.

Probably you could use io.BytesIO which you can initialize with the Response.content. Then instead of list of bytes you get a file-like object.

import io
resp = requests.get(url, stream=True)
obj.mp3 = io.BytesIO(resp.content)

edited Feb 12, 2014 at 1:01

answered Feb 12, 2014 at 0:54

dnll

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1 Comment

eN_Joy Over a year ago

after hours of trying to figure out, i tend to conclude that this is not at all possible: django template expect a string, in particular for any audio player, the string needs to either represent an URL, or a path pointing to the media. however both yours and mine code return an object. thanks anyway.

Sam Dolan · Accepted Answer · 2014-02-12 04:25:03Z

1

Django's responsibility is to generate the HTML code that is then interpreted by the browser. The browser is what needs to be streaming the audio. You need to pass the mp3 url through Django templates that a player like http://www.jwplayer.com/ can then stream on the client side.

answered Feb 12, 2014 at 4:25

Sam Dolan

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