How to send a "multipart/form-data" with requests in python?
Solution 1:
Basically, if you specify a files
parameter (a dictionary), then requests
will send a multipart/form-data
POST instead of a application/x-www-form-urlencoded
POST. You are not limited to using actual files in that dictionary, however:
>>> import requests
>>> response = requests.post('http://httpbin.org/post', files=dict(foo='bar'))
>>> response.status_code
200
and httpbin.org lets you know what headers you posted with; in response.json()
we have:
>>> from pprint import pprint
>>> pprint(response.json()['headers'])
{'Accept': '*/*',
'Accept-Encoding': 'gzip, deflate',
'Connection': 'close',
'Content-Length': '141',
'Content-Type': 'multipart/form-data; '
'boundary=c7cbfdd911b4e720f1dd8f479c50bc7f',
'Host': 'httpbin.org',
'User-Agent': 'python-requests/2.21.0'}
Better still, you can further control the filename, content type and additional headers for each part by using a tuple instead of a single string or bytes object. The tuple is expected to contain between 2 and 4 elements; the filename, the content, optionally a content type, and an optional dictionary of further headers.
I'd use the tuple form with None
as the filename, so that the filename="..."
parameter is dropped from the request for those parts:
>>> files = {'foo': 'bar'}
>>> print(requests.Request('POST', 'http://httpbin.org/post', files=files).prepare().body.decode('utf8'))
--bb3f05a247b43eede27a124ef8b968c5
Content-Disposition: form-data; name="foo"; filename="foo"
bar
--bb3f05a247b43eede27a124ef8b968c5--
>>> files = {'foo': (None, 'bar')}
>>> print(requests.Request('POST', 'http://httpbin.org/post', files=files).prepare().body.decode('utf8'))
--d5ca8c90a869c5ae31f70fa3ddb23c76
Content-Disposition: form-data; name="foo"
bar
--d5ca8c90a869c5ae31f70fa3ddb23c76--
files
can also be a list of two-value tuples, if you need ordering and/or multiple fields with the same name:
requests.post(
'http://requestb.in/xucj9exu',
files=(
('foo', (None, 'bar')),
('foo', (None, 'baz')),
('spam', (None, 'eggs')),
)
)
If you specify both files
and data
, then it depends on the value of data
what will be used to create the POST body. If data
is a string, only it willl be used; otherwise both data
and files
are used, with the elements in data
listed first.
There is also the excellent requests-toolbelt
project, which includes advanced Multipart support. It takes field definitions in the same format as the files
parameter, but unlike requests
, it defaults to not setting a filename parameter. In addition, it can stream the request from open file objects, where requests
will first construct the request body in memory:
from requests_toolbelt.multipart.encoder import MultipartEncoder
mp_encoder = MultipartEncoder(
fields={
'foo': 'bar',
# plain file object, no filename or mime type produces a
# Content-Disposition header with just the part name
'spam': ('spam.txt', open('spam.txt', 'rb'), 'text/plain'),
}
)
r = requests.post(
'http://httpbin.org/post',
data=mp_encoder, # The MultipartEncoder is posted as data, don't use files=...!
# The MultipartEncoder provides the content-type header with the boundary:
headers={'Content-Type': mp_encoder.content_type}
)
Fields follow the same conventions; use a tuple with between 2 and 4 elements to add a filename, part mime-type or extra headers. Unlike the files
parameter, no attempt is made to find a default filename
value if you don't use a tuple.
Solution 2:
Requests has changed since some of the previous answers were written. Have a look at this Issue on Github for more details and this comment for an example.
In short, the files
parameter takes a dictionary with the key being the name of the form field and the value being either a string or a 2, 3 or 4-length tuple, as described in the section POST a Multipart-Encoded File in the Requests quickstart:
>>> url = 'http://httpbin.org/post'
>>> files = {'file': ('report.xls', open('report.xls', 'rb'), 'application/vnd.ms-excel', {'Expires': '0'})}
In the above, the tuple is composed as follows:
(filename, data, content_type, headers)
If the value is just a string, the filename will be the same as the key, as in the following:
>>> files = {'obvius_session_id': '72c2b6f406cdabd578c5fd7598557c52'}
Content-Disposition: form-data; name="obvius_session_id"; filename="obvius_session_id"
Content-Type: application/octet-stream
72c2b6f406cdabd578c5fd7598557c52
If the value is a tuple and the first entry is None
the filename property will not be included:
>>> files = {'obvius_session_id': (None, '72c2b6f406cdabd578c5fd7598557c52')}
Content-Disposition: form-data; name="obvius_session_id"
Content-Type: application/octet-stream
72c2b6f406cdabd578c5fd7598557c52