0

このcurlコマンドをどのように変換しますか:

curl -v -d email=onlinecrapbox@gmail.com -d password=mypassword -X POST https://www.toggl.com/api/v6/sessions.json

urlib2に？

これが機能しない理由:

url=       'https://www.toggl.com/api/v6/sessions.json'
username = 'onlinecrapbox@gmail.com'
password = 'mypassword'

passman = urllib2.HTTPPasswordMgrWithDefaultRealm()
passman.add_password(None, url, username, password)

authhandler = urllib2.HTTPBasicAuthHandler(passman)

opener = urllib2.build_opener(authhandler)

urllib2.install_opener(opener)
pagehandle = urllib2.urlopen(url)

それは私にこのエラーを与えます:

Traceback (most recent call last):
  File "/Users/jorrit/virtualenvs/tiddle/tiddle/troggle/tests.py", line 16, in test_get_troggle_connection
    get_projects()
  File "/Users/jorrit/virtualenvs/tiddle/tiddle/troggle/views.py", line 29, in get_projects
    pagehandle = urllib2.urlopen(url)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 126, in urlopen
    return _opener.open(url, data, timeout)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 400, in open
    response = meth(req, response)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 513, in http_response
    'http', request, response, code, msg, hdrs)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 438, in error
    return self._call_chain(*args)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 372, in _call_chain
    result = func(*args)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 521, in http_error_default
    raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
HTTPError: HTTP Error 403: Forbidden
4

2 に答える 2

1

requests パッケージの場合、コードは次のようになります。

import requests

url = 'https://www.toggl.com/api/v6/sessions.json'
payload = {'email': 'onlinecrapbox@gmail.com',
           'password': 'mypassword'}

r = requests.post(url, data=payload)
于 2012-04-08T13:41:26.117 に答える
0

あなたのurllibコードはcurlコマンド以外のことをするからです。代わりにこれを試してください：

import urllib
import urllib2

url = 'https://www.toggl.com/api/v6/sessions.json'
values = {'email': 'onlinecrapbox@gmail.com',
          'password': 'mypassword'}

data = urllib.urlencode(values)
req = urllib2.Request(url, data)
response = urllib2.urlopen(req)
the_page = response.read()
于 2012-04-08T11:32:46.370 に答える