From wechat api doc:

curl -F [email protected] "http://file.api.wechat.com/cgi-bin/media/upload?access_token=ACCESS_TOKEN&type=TYPE"

Translate the command above to python:

import requests
url = 'http://file.api.wechat.com/cgi-bin/media/upload?access_token=ACCESS_TOKEN&type=TYPE'
files = {'media': open('test.jpg', 'rb')}
requests.post(url, files=files)

Doc: https://docs.python-requests.org/en/master/user/quickstart/#post-a-multipart-encoded-file

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'}

And just to be explicit: you should not set the Content-Type header when you use the files parameter, leave this to requests because it needs to specify a (unique) boundary value in the header that matches the value used in the request body.

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.

Pass in data as a dictionary:

data = {
    'caption', caption,
    'access_token', token
}
files = {
    'file': open(IMG_PATH, 'rb')
}
status = requests.post(
    'https://graph.facebook.com/v2.7/PAGE_ID/photos',
     data=data, files=files)

requests can't produce multipart/form-data parts (together with the files you are uploading) from a application/x-www-form-urlencoded encoded string.

Using a dictionary for the POST data has the additional advantage that requests takes care of properly encoding the values; caption especially could contain data that you must escape properly.

You probably don't need all that much.

Pass the data and files as arguments. The multipart request will be constructed automatically.

See the docs.

import requests

# ...

data = {
    'shipping_sg_other_name': 'atikrahman',
    # .... 
    'shipping_sg_registered_options': 'false',
}

files = ['/path/to/pic.jpg']

response = requests.post(url, data=data, files=files, headers=headers)
print(response.text)

Files are supported in python3 requests module, here. This should work out for you.

import requests

url = "http://xxx"

# just set files to a list of tuples of (form_field_name, file_info)
multiple_files = [
        ('images', ('xxx.jpeg', open('xxx.jpeg', 'rb'), 'image/png')),
        ('images', ('bar.png', open('bar.png', 'rb'), 'image/png'))
]

text_data = {"key":"value"}
headers = {
    "Authorization" : "xxx",
    "Content-Type": "application/json"
}
r = requests.post(url, files=multiple_files, data=text_data, headers=headers)
r.text

try with this

import requests

headers = {
    'Content-Type': 'multipart/form-data',
    'x-api-key': 'xxxxxx-xxxxx-xxxx-xxxx-xxxxxxxx',
}

files = {
    'file': ('<image>.jpg', open('<image>.jpg', 'rb')),
}

response = requests.post('http://localhost:8000/api/v1/recognition/recognize', headers=headers, files=files)

Adding this reply if anyone needs it for future.

Please find the below code for sending image file in Python requests. Here logo is the keyword for an image file, which one of our servers accepts as an input. Hence 'logo' is given. Also, make sure you remove 'Content-type' from the headers.

response = requests.post(url,data=data,headers=headers, files={'logo':open("your_image_file.jpeg",'rb')})