Python HTTP Post fails

Question

After searching for how to POST large files using python, I came across this and I wrote a program based on that, but when I run it I get the following error message:

 Traceback (most recent call last):
 File "Test3.py", line 187, in <module>
 main()
 File "Test3.py", line 184, in main
 do_upload(options, args)
 File "Test3.py", line 48, in do_upload
 response = urllib2.urlopen(request)
 File "C:\Python27\lib\urllib2.py", line 126, in urlopen
 return _opener.open(url, data, timeout)
 File "C:\Python27\lib\urllib2.py", line 400, in open
 response = meth(req, response)
 File "C:\Python27\lib\urllib2.py", line 513, in http_response
 'http', request, response, code, msg, hdrs)
 File "C:\Python27\lib\urllib2.py", line 438, in error
 return self._call_chain(*args)
 File "C:\Python27\lib\urllib2.py", line 372, in _call_chain
 result = func(*args)
 File "C:\Python27\lib\urllib2.py", line 521, in http_error_default
 raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 500: Internal Server Error

and here is my code, to run the program I use --upload "path_to_file" [space] "filename". I'm new to python programming so most of this still confuses me.

def do_upload(options, args):
 url = 'http://127.0.0.1/test_server/upload'
 path = args[0]
 # print path
 filename = args[1]
 if not os.access(args[0], os.F_OK):
 print "Directory/file Doesn't exist"
 exit(1)
 os.chdir(path)
 f = open(filename, 'rb')
 mmapped_file_as_string = mmap.mmap(f.fileno(), 0, access=mmap.ACCESS_READ)
 request = urllib2.Request(url, mmapped_file_as_string)
 contenttype = mimetypes.guess_type(filename)[0] 
 request.add_header(contenttype, 'application/octet-stream')
 response = urllib2.urlopen(request)
 #close everything
 mmapped_file_as_string.close()
 f.close()

UPDATE

I have changed the code from above and now I'm getting some socket error.

Updated code

def do_upload(options, args):
 host = 'http://localhost:80'
 selector = '/test_server/upload'
 url = 'http://localhost:80/test_server/upload'
 if len(args) == 2:
 print "len of args = 2"
 files = "File is " + str(args[1])
 print files
 path = args[0]
 print "Path is " + str(args[0])
 content_type, body = encode_multipart_formdata(files)
 h = httplib.HTTP(host)
 h.putrequest('POST', selector)
 h.putheader('content-type', content_type)
 h.putheader('content-length', str(len(body)))
 h.endheaders()
 h.send(body)
 errcode, errmsg, headers = h.getreply()
 return h.file.read()
 f = open(files, 'rb')
 mmapped_file_as_string = mmap.mmap(f.fileno(), 0, access=mmap.ACCESS_READ)
 request = urllib2.Request(url, mmapped_file_as_string)
 request.add_header('Content-Type', content_type)
 response = urllib2.urlopen(request)
 mmapped_file_as_string.close()
 f.close()
def encode_multipart_formdata(files):
 BOUNDARY = '----------ThIs_Is_tHe_bouNdaRY_$'
 CRLF = '\r\n'
 L = []
 for (filename) in files:
 L.append('--' + BOUNDARY)
 L.append('Content-Disposition: form-data; filename="%s"' % (filename))
 L.append('Content-Type: %s' % get_content_type(filename))
 L.append('')
 #L.append(value)
 L.append('--' + BOUNDARY + '--')
 L.append('')
 body = CRLF.join(L)
 content_type = 'multipart/form-data; boundary=%s' % BOUNDARY
 return content_type, body
def get_content_type(filename):
 return mimetypes.guess_type(filename)[0] or 'application/octet-stream'

Error message

Traceback (most recent call last):
 File "Test3.py", line 208, in <module>
 main()
 File "Test3.py", line 205, in main
 do_upload(options, args)
 File "Test3.py", line 41, in do_upload
 h.endheaders()
 File "C:\Python27\lib\httplib.py", line 951, in endheaders
 self._send_output(message_body)
 File "C:\Python27\lib\httplib.py", line 811, in _send_output
 self.send(msg)
 File "C:\Python27\lib\httplib.py", line 773, in send
 self.connect()
 File "C:\Python27\lib\httplib.py", line 754, in connect
 self.timeout, self.source_address)
 File "C:\Python27\lib\socket.py", line 553, in create_connection
 for res in getaddrinfo(host, port, 0, SOCK_STREAM):
socket.gaierror: [Errno 11004] getaddrinfo failed

• dm03514: can you post test_server/upload code???
• cyberbemon: Done ! , I like to mention one thing , I was able to upload files using cURL with out any problems.

Answer 1

You are setting a non-existing header instead of the Content-Type header:

request.add_header(contenttype, 'application/octet-stream')

Change that to:

request.add_header('Content-Type', contenttype)

instead.

Your biggest problem however is that you are not uploading a multipart POST, but rather just the file itself as the POST body, while your server is expecting only a multipart upload.

Take a look at this SO answer for a very simple way to generate a proper multipart POST body: https://stackoverflow.com/a/681182/100297. Note that you'll have to adjust your Content-Type header accordingly.

• cyberbemon: That Didn't fix it , it changed the first 3 lines from the error messaged and replace them with these File "Test3.py", line 178, in <module> main() File "Test3.py", line 175, in main do_upload(options, args) File "Test3.py", line 39, in do_upload response = urllib2.urlopen(request)
• cyberbemon: Thanks, Looking at the code I'm a little confused as to what def post_multipart(host, selector, fields, files): the selector in that function means
• Martijn Pieters: It's the path portion of the URL; /test_server/upload in your case.
• cyberbemon: So Host = 127.0.0.1 Selector = /test_server/upload Field = Myfile files = Path to file+filename ?
• Martijn Pieters: Your error menas there is no such host ('http://localhost:80'). Read the documentation for HTTP and you'll see you need to provide a bare hostname, so 'localhost'.