upload a file using httplib

I want to share a small snippet of code to upload a file to a remote server as a “multipart/form-data” . The function below gets two arguments. The server url ( ex: http://server.org/upload ) and a filename. First the filename encoded as a “form-data”, then we use httplib to POST it to the server. Since httplib wants the host + path in separate stages, we have to parse the url using urlparse.

The receiving server must accept the data and return the location of the newly created resource. There are many snippet on the web, but I felt they were all incomplete or too messy. The encode function below is actually part of a snippet I found googling around. Happy uploading.

import httplib
import urlparse

def upload(url,filename):
    def encode (file_path, fields=[]):
        BOUNDARY = '----------bundary------'
        CRLF = '\r\n'
        body = []
        # Add the metadata about the upload first
        for key, value in fields:
            body.extend(
              ['--' + BOUNDARY,
               'Content-Disposition: form-data; name="%s"' % key,
               '',
               value,
               ])
        # Now add the file itself
        file_name = os.path.basename(file_path)
        f = open(file_path, 'rb')
        file_content = f.read()
        f.close()
        body.extend(
          ['--' + BOUNDARY,
           'Content-Disposition: form-data; name="file"; filename="%s"'
           % file_name,
           # The upload server determines the mime-type, no need to set it.
           'Content-Type: application/octet-stream',
           '',
           file_content,
           ])
        # Finalize the form body
        body.extend(['--' + BOUNDARY + '--', ''])
        return 'multipart/form-data; boundary=%s' % BOUNDARY, CRLF.join(body)

    if os.path.exists(filename):
        content_type, body = encode(filename)
        headers = { 'Content-Type': content_type }
        u = urlparse.urlparse(url)
        server = httplib.HTTPConnection(u.netloc)
        server.request('POST', u.path, body, headers)
        resp = server.getresponse()
        server.close()

        if resp.status == 201:
            location = resp.getheader('Location', None)
        else :
            print resp.status, resp.reason
            location = None

        return location

Since I’m working with Django, this is the server part. Few remarks: I create the file name using uuid1(). This is an easy way to create unique identifier. A bit over killing maybe. I assume a model myfiles and a form UploadFileForm that you can easily guess. the function handle_uploaded_file is the procedure that actually saves the file on the disk. This is standard. I return a “Location” where the user can access the file. You have to create a small view to serve the file.

import uuid
from django.http import HttpResponse
import os
import datetime
from myapp.models import myfiles
from myapp.forms import UploadFileForm

def handle_uploaded_file(f,n):
    destination = open(n, 'wb+')
    for chunk in f.chunks():
        destination.write(chunk)
    destination.close()

def upload(request):
    if request.method == 'POST':
        form = UploadFileForm(request.POST, request.FILES)
        if form.is_valid():
            ip = request.META['REMOTE_ADDR']
            u = str(uuid.uuid1())
            uploaded = datetime.datetime.now()
            fname = os.path.join(baseupdir, u)
            handle_uploaded_file(request.FILES['file'],fname)
            size = os.path.getsize(fname)

            d = myfiles(fname=fname,size=size,uploaded=uploaded,ip=ip,uuid=u).save()

            response = HttpResponse(content="", status=201)
            response["Location"] = "/file?uuid=%s" % u
            return response # 10.2.2 201 Created
        else :
            return HttpResponse(status=400) # 10.4.1 400 Bad Request
    else :
        return HttpResponse(status=400) # 10.4.1 400 Bad Request