Python: HTTP Post a large file with streaming

Question

I'm uploading potentially large files to a web server. Currently I'm doing this:

import urllib2

f = open('somelargefile.zip','rb')
request = urllib2.Request(url,f.read())
request.add_header("Content-Type", "application/zip")
response = urllib2.urlopen(request)

However, this reads the entire file's contents into memory before posting it. How can I have it stream the file to the server?

Answer 1

Have you tried with Mechanize?

from mechanize import Browser
br = Browser()
br.open(url)
br.form.add_file(open('largefile.zip'), 'application/zip', 'largefile.zip')
br.submit()

or, if you don't want to use multipart/form-data, check this old post.

It suggests two options:

  1. Use mmap, Memory Mapped file object
  2. Patch httplib.HTTPConnection.send

Answer 2

Try pycurl. I don't have anything setup will accept a large file that isn't in a multipart/form-data POST, but here's a simple example that reads the file as needed.

import os
import pycurl

class FileReader:
    def __init__(self, fp):
        self.fp = fp
    def read_callback(self, size):
        return self.fp.read(size)

c = pycurl.Curl()
c.setopt(pycurl.URL, url)
c.setopt(pycurl.UPLOAD, 1)
c.setopt(pycurl.READFUNCTION, FileReader(open(filename, 'rb')).read_callback)
filesize = os.path.getsize(filename)
c.setopt(pycurl.INFILESIZE, filesize)
c.perform()
c.close()

Answer 3

Reading through the mailing list thread linked to by systempuntoout, I found a clue towards the solution.

The mmap module allows you to open file that acts like a string. Parts of the file are loaded into memory on demand.

Here's the code I'm using now:

import urllib2
import mmap

# Open the file as a memory mapped string. Looks like a string, but 
# actually accesses the file behind the scenes. 
f = open('somelargefile.zip','rb')
mmapped_file_as_string = mmap.mmap(f.fileno(), 0, access=mmap.ACCESS_READ)

# Do the request
request = urllib2.Request(url, mmapped_file_as_string)
request.add_header("Content-Type", "application/zip")
response = urllib2.urlopen(request)

#close everything
mmapped_file_as_string.close()
f.close()