Thanks Bartosz...
Trying that,
files = {'file': open(LOCAL_FILENAME,'rb')}
values = {'action': 'upload', 'format':'json','filename':fn,
'comment':'test upload',
'text':'{{OGL2}}\n\nPlease contact [[User:Storkk]] if
there are issues','token':csrftoken}
r = requests.post(baseurl, files=files, data=values)
Doesn't even return JSON, but the HTML API help.
Likewise with
with open(LOCAL_FILENAME,'rb') as f:
CONTENTS = f.read()
files = {'file': CONTENTS}
same thing.
storkk
Quoting Bartosz Dziewoński <[email protected]>:
You should probably just pass the file to the requests library and
have it do all the work, like setting up multipart/form-data and
Content-Disposition. I don't often write Python, but this looks like
the right approach: http://stackoverflow.com/a/22567429
--
Bartosz Dziewoński
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