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I want to send image from my computer to server [duplicate]

I'm performing a simple task of uploading a file using Python requests library. I searched Stack Overflow and no one seemed to have the same problem, namely, that the file is not received by the server:
import requests
url='http://nesssi.cacr.caltech.edu/cgi-bin/getmulticonedb_release2.cgi/post'
files={'files': open('file.txt','rb')}
values={'upload_file' : 'file.txt' , 'DB':'photcat' , 'OUT':'csv' , 'SHORT':'short'}
r=requests.post(url,files=files,data=values)
I'm filling the value of 'upload_file' keyword with my filename, because if I leave it blank, it says
Error - You must select a file to upload!
And now I get
File file.txt of size bytes is uploaded successfully!
Query service results: There were 0 lines.
Which comes up only if the file is empty. So I'm stuck as to how to send my file successfully. I know that the file works because if I go to this website and manually fill in the form it returns a nice list of matched objects, which is what I'm after. I'd really appreciate all hints.
Some other threads related (but not answering my problem):
Send file using POST from a Python script
http://docs.python-requests.org/en/latest/user/quickstart/#response-content
Uploading files using requests and send extra data
http://docs.python-requests.org/en/latest/user/advanced/#body-content-workflow

If upload_file is meant to be the file, use:
files = {'upload_file': open('file.txt','rb')}
values = {'DB': 'photcat', 'OUT': 'csv', 'SHORT': 'short'}
r = requests.post(url, files=files, data=values)
and requests will send a multi-part form POST body with the upload_file field set to the contents of the file.txt file.
The filename will be included in the mime header for the specific field:
>>> import requests
>>> open('file.txt', 'wb') # create an empty demo file
<_io.BufferedWriter name='file.txt'>
>>> files = {'upload_file': open('file.txt', 'rb')}
>>> print(requests.Request('POST', 'http://example.com', files=files).prepare().body.decode('ascii'))
--c226ce13d09842658ffbd31e0563c6bd
Content-Disposition: form-data; name="upload_file"; filename="file.txt"
--c226ce13d09842658ffbd31e0563c6bd--
Note the filename="file.txt" parameter.
You can use a tuple for the files mapping value, with between 2 and 4 elements, if you need more control. The first element is the filename, followed by the contents, and an optional content-type header value and an optional mapping of additional headers:
files = {'upload_file': ('foobar.txt', open('file.txt','rb'), 'text/x-spam')}
This sets an alternative filename and content type, leaving out the optional headers.
If you are meaning the whole POST body to be taken from a file (with no other fields specified), then don't use the files parameter, just post the file directly as data. You then may want to set a Content-Type header too, as none will be set otherwise. See Python requests - POST data from a file.

(2018) the new python requests library has simplified this process, we can use the 'files' variable to signal that we want to upload a multipart-encoded file
url = 'http://httpbin.org/post'
files = {'file': open('report.xls', 'rb')}
r = requests.post(url, files=files)
r.text

Client Upload
If you want to upload a single file with Python requests library, then requests lib supports streaming uploads, which allow you to send large files or streams without reading into memory.
with open('massive-body', 'rb') as f:
requests.post('http://some.url/streamed', data=f)
Server Side
Then store the file on the server.py side such that save the stream into file without loading into the memory. Following is an example with using Flask file uploads.
#app.route("/upload", methods=['POST'])
def upload_file():
from werkzeug.datastructures import FileStorage
FileStorage(request.stream).save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
return 'OK', 200
Or use werkzeug Form Data Parsing as mentioned in a fix for the issue of "large file uploads eating up memory" in order to avoid using memory inefficiently on large files upload (s.t. 22 GiB file in ~60 seconds. Memory usage is constant at about 13 MiB.).
#app.route("/upload", methods=['POST'])
def upload_file():
def custom_stream_factory(total_content_length, filename, content_type, content_length=None):
import tempfile
tmpfile = tempfile.NamedTemporaryFile('wb+', prefix='flaskapp', suffix='.nc')
app.logger.info("start receiving file ... filename => " + str(tmpfile.name))
return tmpfile
import werkzeug, flask
stream, form, files = werkzeug.formparser.parse_form_data(flask.request.environ, stream_factory=custom_stream_factory)
for fil in files.values():
app.logger.info(" ".join(["saved form name", fil.name, "submitted as", fil.filename, "to temporary file", fil.stream.name]))
# Do whatever with stored file at `fil.stream.name`
return 'OK', 200

You can send any file via post api while calling the API just need to mention files={'any_key': fobj}
import requests
import json
url = "https://request-url.com"
headers = {"Content-Type": "application/json; charset=utf-8"}
with open(filepath, 'rb') as fobj:
response = requests.post(url, headers=headers, files={'file': fobj})
print("Status Code", response.status_code)
print("JSON Response ", response.json())

#martijn-pieters answer is correct, however I wanted to add a bit of context to data= and also to the other side, in the Flask server, in the case where you are trying to upload files and a JSON.
From the request side, this works as Martijn describes:
files = {'upload_file': open('file.txt','rb')}
values = {'DB': 'photcat', 'OUT': 'csv', 'SHORT': 'short'}
r = requests.post(url, files=files, data=values)
However, on the Flask side (the receiving webserver on the other side of this POST), I had to use form
#app.route("/sftp-upload", methods=["POST"])
def upload_file():
if request.method == "POST":
# the mimetype here isnt application/json
# see here: https://stackoverflow.com/questions/20001229/how-to-get-posted-json-in-flask
body = request.form
print(body) # <- immutable dict
body = request.get_json() will return nothing. body = request.get_data() will return a blob containing lots of things like the filename etc.
Here's the bad part: on the client side, changing data={} to json={} results in this server not being able to read the KV pairs! As in, this will result in a {} body above:
r = requests.post(url, files=files, json=values). # No!
This is bad because the server does not have control over how the user formats the request; and json= is going to be the habbit of requests users.

Upload:
with open('file.txt', 'rb') as f:
files = {'upload_file': f.read()}
values = {'DB': 'photcat', 'OUT': 'csv', 'SHORT': 'short'}
r = requests.post(url, files=files, data=values)
Download (Django):
with open('file.txt', 'wb') as f:
f.write(request.FILES['upload_file'].file.read())

Regarding the answers given so far, there was always something missing that prevented it to work on my side. So let me show you what worked for me:
import json
import os
import requests
API_ENDPOINT = "http://localhost:80"
access_token = "sdfJHKsdfjJKHKJsdfJKHJKysdfJKHsdfJKHs" # TODO: get fresh Token here
def upload_engagement_file(filepath):
url = API_ENDPOINT + "/api/files" # add any URL parameters if needed
hdr = {"Authorization": "Bearer %s" % access_token}
with open(filepath, "rb") as fobj:
file_obj = fobj.read()
file_basename = os.path.basename(filepath)
file_to_upload = {"file": (str(file_basename), file_obj)}
finfo = {"fullPath": filepath}
upload_response = requests.post(url, headers=hdr, files=file_to_upload, data=finfo)
fobj.close()
# print("Status Code ", upload_response.status_code)
# print("JSON Response ", upload_response.json())
return upload_response
Note that requests.post(...) needs
a url parameter, containing the full URL of the API endpoint you're calling, using the API_ENDPOINT, assuming we have an http://localhost:8000/api/files endpoint to POST a file
a headers parameter, containing at least the authorization (bearer token)
a files parameter taking the name of the file plus the entire file content
a data parameter taking just the path and file name
Installation required (console):
pip install requests
What you get back from the function call is a response object containing a status code and also the full error message in JSON format. The commented print statements at the end of upload_engagement_file are showing you how you can access them.
Note: Some useful additional information about the requests library can be found here

Some may need to upload via a put request and this is slightly different that posting data. It is important to understand how the server expects the data in order to form a valid request. A frequent source of confusion is sending multipart-form data when it isn't accepted. This example uses basic auth and updates an image via a put request.
url = 'foobar.com/api/image-1'
basic = requests.auth.HTTPBasicAuth('someuser', 'password123')
# Setting the appropriate header is important and will vary based
# on what you upload
headers = {'Content-Type': 'image/png'}
with open('image-1.png', 'rb') as img_1:
r = requests.put(url, auth=basic, data=img_1, headers=headers)
While the requests library makes working with http requests a lot easier, some of its magic and convenience obscures just how to craft more nuanced requests.

In Ubuntu you can apply this way,
to save file at some location (temporary) and then open and send it to API
path = default_storage.save('static/tmp/' + f1.name, ContentFile(f1.read()))
path12 = os.path.join(os.getcwd(), "static/tmp/" + f1.name)
data={} #can be anything u want to pass along with File
file1 = open(path12, 'rb')
header = {"Content-Disposition": "attachment; filename=" + f1.name, "Authorization": "JWT " + token}
res= requests.post(url,data,header)

how to upload a file using request.Post in python [duplicate]

I'm performing a simple task of uploading a file using Python requests library. I searched Stack Overflow and no one seemed to have the same problem, namely, that the file is not received by the server:
import requests
url='http://nesssi.cacr.caltech.edu/cgi-bin/getmulticonedb_release2.cgi/post'
files={'files': open('file.txt','rb')}
values={'upload_file' : 'file.txt' , 'DB':'photcat' , 'OUT':'csv' , 'SHORT':'short'}
r=requests.post(url,files=files,data=values)
I'm filling the value of 'upload_file' keyword with my filename, because if I leave it blank, it says
Error - You must select a file to upload!
And now I get
File file.txt of size bytes is uploaded successfully!
Query service results: There were 0 lines.
Which comes up only if the file is empty. So I'm stuck as to how to send my file successfully. I know that the file works because if I go to this website and manually fill in the form it returns a nice list of matched objects, which is what I'm after. I'd really appreciate all hints.
Some other threads related (but not answering my problem):
Send file using POST from a Python script
http://docs.python-requests.org/en/latest/user/quickstart/#response-content
Uploading files using requests and send extra data
http://docs.python-requests.org/en/latest/user/advanced/#body-content-workflow

If upload_file is meant to be the file, use:
files = {'upload_file': open('file.txt','rb')}
values = {'DB': 'photcat', 'OUT': 'csv', 'SHORT': 'short'}
r = requests.post(url, files=files, data=values)
and requests will send a multi-part form POST body with the upload_file field set to the contents of the file.txt file.
The filename will be included in the mime header for the specific field:
>>> import requests
>>> open('file.txt', 'wb') # create an empty demo file
<_io.BufferedWriter name='file.txt'>
>>> files = {'upload_file': open('file.txt', 'rb')}
>>> print(requests.Request('POST', 'http://example.com', files=files).prepare().body.decode('ascii'))
--c226ce13d09842658ffbd31e0563c6bd
Content-Disposition: form-data; name="upload_file"; filename="file.txt"
--c226ce13d09842658ffbd31e0563c6bd--
Note the filename="file.txt" parameter.
You can use a tuple for the files mapping value, with between 2 and 4 elements, if you need more control. The first element is the filename, followed by the contents, and an optional content-type header value and an optional mapping of additional headers:
files = {'upload_file': ('foobar.txt', open('file.txt','rb'), 'text/x-spam')}
This sets an alternative filename and content type, leaving out the optional headers.
If you are meaning the whole POST body to be taken from a file (with no other fields specified), then don't use the files parameter, just post the file directly as data. You then may want to set a Content-Type header too, as none will be set otherwise. See Python requests - POST data from a file.

(2018) the new python requests library has simplified this process, we can use the 'files' variable to signal that we want to upload a multipart-encoded file
url = 'http://httpbin.org/post'
files = {'file': open('report.xls', 'rb')}
r = requests.post(url, files=files)
r.text

Client Upload
If you want to upload a single file with Python requests library, then requests lib supports streaming uploads, which allow you to send large files or streams without reading into memory.
with open('massive-body', 'rb') as f:
requests.post('http://some.url/streamed', data=f)
Server Side
Then store the file on the server.py side such that save the stream into file without loading into the memory. Following is an example with using Flask file uploads.
#app.route("/upload", methods=['POST'])
def upload_file():
from werkzeug.datastructures import FileStorage
FileStorage(request.stream).save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
return 'OK', 200
Or use werkzeug Form Data Parsing as mentioned in a fix for the issue of "large file uploads eating up memory" in order to avoid using memory inefficiently on large files upload (s.t. 22 GiB file in ~60 seconds. Memory usage is constant at about 13 MiB.).
#app.route("/upload", methods=['POST'])
def upload_file():
def custom_stream_factory(total_content_length, filename, content_type, content_length=None):
import tempfile
tmpfile = tempfile.NamedTemporaryFile('wb+', prefix='flaskapp', suffix='.nc')
app.logger.info("start receiving file ... filename => " + str(tmpfile.name))
return tmpfile
import werkzeug, flask
stream, form, files = werkzeug.formparser.parse_form_data(flask.request.environ, stream_factory=custom_stream_factory)
for fil in files.values():
app.logger.info(" ".join(["saved form name", fil.name, "submitted as", fil.filename, "to temporary file", fil.stream.name]))
# Do whatever with stored file at `fil.stream.name`
return 'OK', 200

You can send any file via post api while calling the API just need to mention files={'any_key': fobj}
import requests
import json
url = "https://request-url.com"
headers = {"Content-Type": "application/json; charset=utf-8"}
with open(filepath, 'rb') as fobj:
response = requests.post(url, headers=headers, files={'file': fobj})
print("Status Code", response.status_code)
print("JSON Response ", response.json())

#martijn-pieters answer is correct, however I wanted to add a bit of context to data= and also to the other side, in the Flask server, in the case where you are trying to upload files and a JSON.
From the request side, this works as Martijn describes:
files = {'upload_file': open('file.txt','rb')}
values = {'DB': 'photcat', 'OUT': 'csv', 'SHORT': 'short'}
r = requests.post(url, files=files, data=values)
However, on the Flask side (the receiving webserver on the other side of this POST), I had to use form
#app.route("/sftp-upload", methods=["POST"])
def upload_file():
if request.method == "POST":
# the mimetype here isnt application/json
# see here: https://stackoverflow.com/questions/20001229/how-to-get-posted-json-in-flask
body = request.form
print(body) # <- immutable dict
body = request.get_json() will return nothing. body = request.get_data() will return a blob containing lots of things like the filename etc.
Here's the bad part: on the client side, changing data={} to json={} results in this server not being able to read the KV pairs! As in, this will result in a {} body above:
r = requests.post(url, files=files, json=values). # No!
This is bad because the server does not have control over how the user formats the request; and json= is going to be the habbit of requests users.

Upload:
with open('file.txt', 'rb') as f:
files = {'upload_file': f.read()}
values = {'DB': 'photcat', 'OUT': 'csv', 'SHORT': 'short'}
r = requests.post(url, files=files, data=values)
Download (Django):
with open('file.txt', 'wb') as f:
f.write(request.FILES['upload_file'].file.read())

Regarding the answers given so far, there was always something missing that prevented it to work on my side. So let me show you what worked for me:
import json
import os
import requests
API_ENDPOINT = "http://localhost:80"
access_token = "sdfJHKsdfjJKHKJsdfJKHJKysdfJKHsdfJKHs" # TODO: get fresh Token here
def upload_engagement_file(filepath):
url = API_ENDPOINT + "/api/files" # add any URL parameters if needed
hdr = {"Authorization": "Bearer %s" % access_token}
with open(filepath, "rb") as fobj:
file_obj = fobj.read()
file_basename = os.path.basename(filepath)
file_to_upload = {"file": (str(file_basename), file_obj)}
finfo = {"fullPath": filepath}
upload_response = requests.post(url, headers=hdr, files=file_to_upload, data=finfo)
fobj.close()
# print("Status Code ", upload_response.status_code)
# print("JSON Response ", upload_response.json())
return upload_response
Note that requests.post(...) needs
a url parameter, containing the full URL of the API endpoint you're calling, using the API_ENDPOINT, assuming we have an http://localhost:8000/api/files endpoint to POST a file
a headers parameter, containing at least the authorization (bearer token)
a files parameter taking the name of the file plus the entire file content
a data parameter taking just the path and file name
Installation required (console):
pip install requests
What you get back from the function call is a response object containing a status code and also the full error message in JSON format. The commented print statements at the end of upload_engagement_file are showing you how you can access them.
Note: Some useful additional information about the requests library can be found here

Some may need to upload via a put request and this is slightly different that posting data. It is important to understand how the server expects the data in order to form a valid request. A frequent source of confusion is sending multipart-form data when it isn't accepted. This example uses basic auth and updates an image via a put request.
url = 'foobar.com/api/image-1'
basic = requests.auth.HTTPBasicAuth('someuser', 'password123')
# Setting the appropriate header is important and will vary based
# on what you upload
headers = {'Content-Type': 'image/png'}
with open('image-1.png', 'rb') as img_1:
r = requests.put(url, auth=basic, data=img_1, headers=headers)
While the requests library makes working with http requests a lot easier, some of its magic and convenience obscures just how to craft more nuanced requests.

In Ubuntu you can apply this way,
to save file at some location (temporary) and then open and send it to API
path = default_storage.save('static/tmp/' + f1.name, ContentFile(f1.read()))
path12 = os.path.join(os.getcwd(), "static/tmp/" + f1.name)
data={} #can be anything u want to pass along with File
file1 = open(path12, 'rb')
header = {"Content-Disposition": "attachment; filename=" + f1.name, "Authorization": "JWT " + token}
res= requests.post(url,data,header)

Python 2.7 requests REST API call not working same as curl

I am trying to do a curl post query using requests python 2.7, however the API response differently using curl vs. requests lib.
The post query is pretty simple a file and name-value-pair data as the API params.
Below is the curl multipart post request:
curl -uadmin:blabla123 -X POST 127.0.0.1:8080/alfresco/api/-default-/public/alfresco/versions/1/nodes/6a0ab661-1c43-43ed-b07f-a564f6bcb5ca/children -F filedata=#file1.txt -F name=document__55;nodeType=content
The python 2.7 code is as below:
import requests
from requests.auth import HTTPBasicAuth, HTTPDigestAuth
from config import USER, PASSWD
def createDocument( documentFilename, documentMetadata, targetFolderNodeId):
'''
Uploads a file and its meta-data to the CMIS server under the specified
target folder
'''
with open(documentFilename, 'rb') as file:
files = {'file': file}
# createURL = 'http://127.0.0.1:8080/alfresco/api/-default-/public/alfresco/versions/1/nodes/{0}/children'.format( targetFolderNodeId )
createURL = 'http://127.0.0.1:8080/alfresco/api/-default-/public/alfresco/versions/1/nodes/6a0ab661-1c43-43ed-b07f-a564f6bcb5ca/children'
data = {
"name":"document__55",
"nodeType":"cm:content",
}
response = requests.post( createURL, data = data, files = files, auth=HTTPBasicAuth(USER, PASSWD) )
print(response)
print(response.json)
print(response.text)
createDocument('file1.txt', '', '')
Curl returns 200 http code but the script oddly returns 400.
Any help is much appreciate it.

At a first glace the reasons for that script failing vs curl working could be more than one:
nodeType=content in the curl, while "nodeType":"cm:content" in the script
there's an additional comma in the script's payload second line
you're not setting the Content-Type header as multipart/form-data in the script, which curl's -F option actually does
Also, the endpoint should provide a verbose error along with the 400 response. If it does, that could be helpful to identify the error cause.

in fact in the curl I have -F filedata=#file1.txt but in the script I have files = {'file': file} which is not the same so I just had to use files = {'filedata': file} :)
def createDocument( documentFilename, documentMetadata, targetFolderNodeId):
'''
Uploads a file and its meta-data to the CMIS server under the specified
target folder
'''
with open(documentFilename, 'rb') as file:
files = {'filedata': file}
createURL = 'http://127.0.0.1:8080/alfresco/api/-default-/public/alfresco/versions/1/nodes/{0}/children'.format( targetFolderNodeId )
data = {
"name":"document__77",
"nodeType":"cm:content"
}
response = requests.post( createURL, data = data, files = files, auth=HTTPBasicAuth(USER, PASSWD) )
print(response)
print(response.json)
print(response.text)
print(response.headers)
# print([i for i in dir(response) if 'header' in i])

Simulate multipart/form-data file upload with Falcon's Testing module

This simple Falcon API will take a HTTP POST with enctype=multipart/form-data and a file upload in the file parameter and print the file's content on the console:
# simple_api.py
import cgi
import falcon
class SomeTestApi(object):
def on_post(self, req, resp):
upload = cgi.FieldStorage(fp=req.stream, environ=req.env)
upload = upload['file'].file.read()
print(upload)
app = falcon.API()
app.add_route('/', SomeTestApi())
One might also use the falcon-multipart middleware to achieve the same goal.
To try it out, run it e.g. with gunicorn (pip install gunicorn),
gunicorn simple_api.py
then use cUrl (or any REST client of choice) to upload a text file:
# sample.txt
this is some sample text
curl -F "file=#sample.txt" localhost:8000
I would like to test this API now with Falcon's testing helpers by simulating a file upload. However, I do not understand yet how to do this (if it is possible at all?). The simulate_request method has a file_wrapper parameter which might be useful but from the documentation I do not understand how this is supposed to be filled.
Any suggestions?

This is what I came up with, which tries to simulate what my Chrome does.
Note that this simulates the case when you are uploading only one file, but you can simply modify this function to upload multiple files, each one separated by two new lines.
def create_multipart(data, fieldname, filename, content_type):
"""
Basic emulation of a browser's multipart file upload
"""
boundry = '----WebKitFormBoundary' + random_string(16)
buff = io.BytesIO()
buff.write(b'--')
buff.write(boundry.encode())
buff.write(b'\r\n')
buff.write(('Content-Disposition: form-data; name="%s"; filename="%s"' % \
(fieldname, filename)).encode())
buff.write(b'\r\n')
buff.write(('Content-Type: %s' % content_type).encode())
buff.write(b'\r\n')
buff.write(b'\r\n')
buff.write(data)
buff.write(b'\r\n')
buff.write(boundry.encode())
buff.write(b'--\r\n')
headers = {'Content-Type': 'multipart/form-data; boundary=%s' %boundry}
headers['Content-Length'] = str(buff.tell())
return buff.getvalue(), headers
You can then use this function like the following:
with open('test/resources/foo.pdf', 'rb') as f:
foodata = f.read()
# Create the multipart data
data, headers = create_multipart(foodata, fieldname='uploadFile',
filename='foo.pdf',
content_type='application/pdf')
# Post to endpoint
client.simulate_request(method='POST', path=url,
headers=headers, body=data)

You can craft a suitable request body and Content-Type using the encode_multipart_formdata function in urllib3, documented here. An example usage:
from falcon import testing
import pytest
import myapp
import urllib3
# Depending on your testing strategy and how your application
# manages state, you may be able to broaden the fixture scope
# beyond the default 'function' scope used in this example.
#pytest.fixture()
def client():
# Assume the hypothetical `myapp` package has a function called
# `create()` to initialize and return a `falcon.App` instance.
return testing.TestClient(myapp.create())
# a dictionary mapping the HTML form label to the file uploads
fields = {
'file_1_form_label': ( # label in HTML form object
'file1.txt', # filename
open('path/to/file1.txt').read(), # file contents
'text/plain' # MIME type
),
'file_2_form_label': (
'file2.json',
open('path/to/file2.json').read(),
'application/json'
)
}
# create the body and header
body, content_type_header = urllib3.encode_multipart_formdata(fields)
# NOTE: modify these headers to reflect those generated by your browser
# and/or required by the falcon application you're testing
headers = {
'Content-Type': content_type_header,
}
# craft the mock query using the falcon testing framework
response = client.simulate_request(
method="POST",
path='/app_path',
headers=headers,
body=body)
print(response.status_code)
Note the syntax of the fields object, which is used as input for the encode_multipart_formdata function.
See Tim Head's blog post for another example:
https://betatim.github.io/posts/python-create-multipart-formdata/
Falcon testing example copied from their docs:
https://falcon.readthedocs.io/en/stable/api/testing.html

Python 3 script to upload a file to a REST URL (multipart request)

I am fairly new to Python and using Python 3.2. I am trying to write a python script that will pick a file from user machine (such as an image file) and submit it to a server using REST based invocation. The Python script should invoke a REST URL and submit the file when the script is called.
This is similar to multipart POST that is done by browser when uploading a file; but here I want to do it through Python script.
If possible do not want to add any external libraries to Python and would like to keep it fairly simple python script using the core Python install.
Can some one guide me? or share some script example that achieve what I want?

Requests library is what you need. You can install with pip install requests.
http://docs.python-requests.org/en/latest/user/quickstart/#post-a-multipart-encoded-file
>>> url = 'http://httpbin.org/post'
>>> files = {'file': open('report.xls', 'rb')}
>>> r = requests.post(url, files=files)

A RESTful way to upload an image would be to use PUT request if you know what the image url is:
#!/usr/bin/env python3
import http.client
h = http.client.HTTPConnection('example.com')
h.request('PUT', '/file/pic.jpg', open('pic.jpg', 'rb'))
print(h.getresponse().read())
upload_docs.py contains an example how to upload a file as multipart/form-data with basic http authentication. It supports both Python 2.x and Python 3.
You could use also requests to post files as a multipart/form-data:
#!/usr/bin/env python3
import requests
response = requests.post('http://httpbin.org/post',
files={'file': open('filename','rb')})
print(response.content)

You can also use unirest . Sample code
import unirest
# consume async post request
def consumePOSTRequestSync():
params = {'test1':'param1','test2':'param2'}
# we need to pass a dummy variable which is open method
# actually unirest does not provide variable to shift between
# application-x-www-form-urlencoded and
# multipart/form-data
params['dummy'] = open('dummy.txt', 'r')
url = 'http://httpbin.org/post'
headers = {"Accept": "application/json"}
# call get service with headers and params
response = unirest.post(url, headers = headers,params = params)
print "code:"+ str(response.code)
print "******************"
print "headers:"+ str(response.headers)
print "******************"
print "body:"+ str(response.body)
print "******************"
print "raw_body:"+ str(response.raw_body)
# post sync request multipart/form-data
consumePOSTRequestSync()
You can check out this post http://stackandqueue.com/?p=57 for more details