python urllib2 file send problem

Question

I want to post a file to a server via python, for this I need to name this file as "xmlfile" so that server recognizes the input.

import urllib2
url = "http://somedomain"
to_send = open('test.xml').read()
data = {}
data['xmlfile'] = to_send
f = urllib2.urlopen(url, data)

This doesn't work, in addition, how can I retrieve the response and save someplace ?

In other words, I want to do the action as I do with Curl:

curl.exe http://somedomain -F [email protected] -o response.html

• Lennart Regebro: possible duplicate of Using MultipartPostHandler to POST form-data with Python

Answer 1

I just read the question nimrodm referred to. One answer mentions the poster module. This module can do the multipart/form-data encoding, so if adding another dependency to your project is not a problem I would go with the poster module.

---

This is not as simple as it should be. There is a code snippet floating around the net which I used for my code and it does the trick. You will probably have to adapt it for your needs.

class RequestWithMethod(urllib2.Request):
 def __init__(self, method, *args, **kwargs):
 self._method = method
 urllib2.Request.__init__(self, *args, **kwargs)
 def get_method(self):
 return self._method
class RestRequest(object):
 def __init__(self, base_url):
 self.base_url = base_url
 def request(self, url, method, headers={"Accept" : "application/json"}, data=None, json_response=True):
 request = RequestWithMethod(url='{0}{1}{2}'.format(self.base_url, root_url(), url),
 method=method,
 headers=headers)
 if data != None:
 data = urllib.urlencode(data)
 response = urllib2.urlopen(request, data=data).read()
 if json_response:
 return from_json(response)
 else:
 return response
 def GET(self, url, **kwargs):
 return self.request(url, 'GET', **kwargs)
 def POST(self, url, **kwargs):
 return self.request(url, 'POST', **kwargs)
 def POST_FILE(self, url, file, headers={"Accept" : "application/json"}, data={}, **kwargs):
 content_type, body = encode_multipart_formdata(data, file)
 headers['Content-type'] = content_type
 headers['Content-length'] = str(len(body))
 request = RequestWithMethod(url='{0}{1}{2}'.format(self.base_url, root_url(), url),
 data=body,
 method='POST',
 headers=headers)
 return from_json(urllib2.urlopen(request).read())
 def PUT(self, url, **kwargs):
 return self.request(url, 'PUT', **kwargs)
 def DELETE(self, url, **kwargs):
 return self.request(url, 'DELETE', **kwargs)
def encode_multipart_formdata(data, file):
 boundary = '----------ThIs_Is_tHe_bouNdaRY_$'
 L = []
 for key, value in data.items():
 L.append('--' + boundary)
 L.append('Content-Disposition: form-data; name="{0}"'.format(key))
 L.append('')
 L.append(value)
 key, filename, value = file
 L.append('--' + boundary)
 L.append('Content-Disposition: form-data; name="{0}"; filename="{1}"'.format(key, filename))
 content_type = mimetypes.guess_type(filename)[0] or 'application/octet-stream'
 L.append('Content-Type: {0}'.format(content_type))
 L.append('')
 L.append(value)
 L.append('--' + boundary + '--')
 L.append('')
 body = '\r\n'.join(L)
 content_type = 'multipart/form-data; boundary={0}'.format(boundary)
 return content_type, body

Answer 2

Look at this stackoverflow question or directly at the code it references.

urllib normally sends data using application/x-www-form-urlencoded content-type and while you need multipart/form-data. The referenced library encodes the data as required.