2011-10-09 111 views
0

我正在使用Web2Py创建一个简单的应用程序,通过UrbanAhiphip发送Push通知。出于某种原因,当我尝试通过我的代码发送时,我收到了400条回应。它使用REST客户端的UA API工作正常。这是我的代码:发送POST请求到UrbanAirship时的HTTP 400响应

url = 'https://go.urbanairship.com/api/push/' 

passman = urllib2.HTTPPasswordMgrWithDefaultRealm() 
# this creates a password manager 

passman.add_password(None, url, username, password) 
# because we have put None at the start it will always 
# use this username/password combination for urls 
# for which `theurl` is a super-url 

authhandler = urllib2.HTTPBasicAuthHandler(passman) 
# create the AuthHandler 

opener = urllib2.build_opener(authhandler) 

urllib2.install_opener(opener) 
# All calls to urllib2.urlopen will now use our handler 
# Make sure not to include the protocol in with the URL, or 
# HTTPPasswordMgrWithDefaultRealm will be very confused. 
# You must (of course) use it when fetching the page though. 

values = {"device_tokens": ["<DEVICE TOKEN>"], "aps": {"alert": "Hello!"}} 

data = urllib.urlencode(values) 
headers = {'Content-Type': 'application/json'} 

req = urllib2.Request(url, data, headers) 

try: 
    response = urllib2.urlopen(req) 
    return response 
except IOError, e: 
    if e.code == 200: 
     return "Push sent!" 
    else: 
     return 'The server couldn\'t fulfill the request. Error: %d' % e.code 

据我可以理解,问题是数据的格式发送。我哪里错了?

回答

0

urllib.urlencode函数用于制作URL编码参数体(Content-Type: application/x-www-form-urlencoded)。对于显然是您想要的JSON,请改用json.dumps

+0

我完全错过了这一点,谢谢! – Mus