Python Requests Multipart HTTP POST

Question

I was wondering how do you translate something like this using Python Requests? In urllib2, you can manually manipulate the data that is being sent over the wire to the API

Answer 1

with requests, I believe that you don't have to be so manual, simply:

import requests

# ...
url = self._resolve_url('/a/creative/uploadcreative')
files = {'file': ('userfile', open(filepath, 'rb'))}
data = {'account_id': account_id}
headers = {'content-type': 'multipart/form-data'}
res = requests.post(url, files=files, data=data, headers=headers)
return res.json

I suppose your concern lies with your:

parts.append('Content-Type: %s' % mimetypes.guess_type(file_path)[0] or 'application/octet-stream')

I haven't proven it to myself beyond the shadow of a doubt. But, I think that is built in to requests here.

Edit: It looks like you can have the normal fields in the files dict, as you propose:

files = {'file': open('image.jpg', 'rb'), 'account_id': 12345}

and could name the filename as you want:

files = {'file': ('userfile', open('image.jpg', 'rb')), 'account_id': 12345}

but, you would get a body.write(b'Content-Type: text/plain\r\n\r\n') on the account_id field which is probably not what you want and have no way to customize the Content-Disposition for each field (still not sure why you would need to); for both the file and the field you will get: Content-Disposition: form-data - which is what you show for both.

I'm not sure that you can do exactly what you want with requests, maybe you should try a feature request.

Answer 2

I found out that in the python-requests library (v.0.13.3), your data will get wiped if you include the "data" field before the "files" field in the request call itself.

For example,

requests.post(url, headers=headers, data=data, files=files) 

will yield empty form-data. However, the following will send the data dictionary as form-data

requests.post(url, headers=headers, files=files, data=data)

Thanks everyone for their answers

Answer 3

 import requests

 import urllib

 def upload_creative(self, account_id, file_path):

    files = [('userfile', (file_path, open(file_path, 'rb'), "image/jpeg" ))]

    url = self._resolve_url('/a/creative/uploadcreative')

    url =  url + "?"  +  urlib.urlencode(account_id=account_id)

    reuests.post(url, files=files)