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Encode multipart form data for uploading files via POST (Python recipe) by Ben Hoyt
ActiveState Code (http://code.activestate.com/recipes/578668/)

This function lets you encode form fields and files in multipart/form-data format for uploading files via HTTP POST.

Python, 82 lines
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"""Encode multipart form data to upload files via POST."""

from __future__ import print_function

import mimetypes
import random
import string

_BOUNDARY_CHARS = string.digits + string.ascii_letters

def encode_multipart(fields, files, boundary=None):
    r"""Encode dict of form fields and dict of files as multipart/form-data.
    Return tuple of (body_string, headers_dict). Each value in files is a dict
    with required keys 'filename' and 'content', and optional 'mimetype' (if
    not specified, tries to guess mime type or uses 'application/octet-stream').

    >>> body, headers = encode_multipart({'FIELD': 'VALUE'},
    ...                                  {'FILE': {'filename': 'F.TXT', 'content': 'CONTENT'}},
    ...                                  boundary='BOUNDARY')
    >>> print('\n'.join(repr(l) for l in body.split('\r\n')))
    '--BOUNDARY'
    'Content-Disposition: form-data; name="FIELD"'
    ''
    'VALUE'
    '--BOUNDARY'
    'Content-Disposition: form-data; name="FILE"; filename="F.TXT"'
    'Content-Type: text/plain'
    ''
    'CONTENT'
    '--BOUNDARY--'
    ''
    >>> print(sorted(headers.items()))
    [('Content-Length', '193'), ('Content-Type', 'multipart/form-data; boundary=BOUNDARY')]
    >>> len(body)
    193
    """
    def escape_quote(s):
        return s.replace('"', '\\"')

    if boundary is None:
        boundary = ''.join(random.choice(_BOUNDARY_CHARS) for i in range(30))
    lines = []

    for name, value in fields.items():
        lines.extend((
            '--{0}'.format(boundary),
            'Content-Disposition: form-data; name="{0}"'.format(escape_quote(name)),
            '',
            str(value),
        ))

    for name, value in files.items():
        filename = value['filename']
        if 'mimetype' in value:
            mimetype = value['mimetype']
        else:
            mimetype = mimetypes.guess_type(filename)[0] or 'application/octet-stream'
        lines.extend((
            '--{0}'.format(boundary),
            'Content-Disposition: form-data; name="{0}"; filename="{1}"'.format(
                    escape_quote(name), escape_quote(filename)),
            'Content-Type: {0}'.format(mimetype),
            '',
            value['content'],
        ))

    lines.extend((
        '--{0}--'.format(boundary),
        '',
    ))
    body = '\r\n'.join(lines)

    headers = {
        'Content-Type': 'multipart/form-data; boundary={0}'.format(boundary),
        'Content-Length': str(len(body)),
    }

    return (body, headers)

if __name__ == '__main__':
    import doctest
    doctest.testmod()
Tags: form, http, multipart, upload

2 comments

Ben Hoyt (author) 10 years, 7 months ago  # | flag

An example of how to use this with Python's built-in urllib2:

import urllib2

import formdata

fields = {'name': 'BOB SMITH'}
files = {'file': {'filename': 'F.DAT', 'content': 'DATA HERE'}}
data, headers = formdata.encode_multipart(fields, files)
request = urllib2.Request('http://httpbin.org/post', data=data, headers=headers)
f = urllib2.urlopen(request)
print f.read()

And the response:

{
  "origin": "<ip address>",
  "json": null,
  "form": {
    "name": "BOB SMITH"
  },
  "url": "http://httpbin.org/post",
  "args": {},
  "headers": {
    "Accept-Encoding": "identity",
    "User-Agent": "Python-urllib/2.7",
    "Content-Length": "278",
    "Connection": "close",
    "Host": "httpbin.org",
    "Content-Type": "multipart/form-data; boundary=nwxUuePw4tNxnJqfcLQem2PLZJFBQS"
  },
  "files": {
    "file": "DATA HERE"
  },
  "data": ""
}
rho 6 years, 11 months ago  # | flag

thanx finally understood!