python urllib2 file send problem

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 = "开发者_Go百科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 xmlfile=@test.xml -o response.html

---

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

---

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.