supply a filename for a file-like object created by urlopen() or requests.get()

问题

I am building a Telegram bot using a Telepot library. To send a picture downloaded from Internet I have to use sendPhoto method, which accepts a file-like object.

Looking through the docs I found this advice:

If the file-like object is obtained by urlopen(), you most likely have to supply a filename because Telegram servers require to know the file extension.

So the question is, if I get my filelike object by opening it with requests.get and wrapping with BytesIO like so:

res = requests.get(some_url)
tbot.sendPhoto(
    messenger_id,
    io.BytesIO(res.content)
)

how and where do I supply a filename?

回答1:

You would supply the filename as the object's .name attribute.

Opening a file with open() has a .name attribute.

>>>local_file = open("file.txt")
>>>local_file
<open file 'file.txt', mode 'r' at ADDRESS>
>>>local_file.name
'file.txt'

Where opening a url does not. Which is why the documentation specifically mentions this.

>>>import urllib
>>>url_file = urllib.open("http://example.com/index.hmtl")
>>>url_file
<addinfourl at 44 whose fp = <open file 'nul', mode 'rb' at ADDRESS>>
>>>url_file.name
AttributeError: addinfourl instance has no attribute 'name'

In your case, you would need to create the file-like object, and give it a .name attribute:

res = requests.get(some_url)
the_file = io.BytesIO(res.content)
the_file.name = 'file.image'

tbot.sendPhoto(
    messenger_id,
    the_file
)

来源：https://stackoverflow.com/questions/42809451/supply-a-filename-for-a-file-like-object-created-by-urlopen-or-requests-get