2010-07-27 70 views
0

我遵循了http://code.google.com/p/geopy/wiki/GettingStarted 教程这工作得很好:Geopy在Django:JSONDecodeError

g = geocoders.Google(resource='maps') 

我想使用JSON作为输出格式,因为我要处理JavaScript中的结果。

,但每次我用:

g = geocoders.Google(resource='maps', output_format='json') 

我得到的错误如下:

Environment: 

Request Method: GET 
Request URL: http://localhost:8000/business/view/13/ 
Django Version: 1.2.1 
Python Version: 2.5.1 
Installed Applications: 
['django.contrib.auth', 
'django.contrib.contenttypes', 
'django.contrib.sessions', 
'django.contrib.sites', 
'django.contrib.messages', 
'registration', 
'pley.business', 
'pley.review', 
'django.contrib.admin', 
'django.contrib.webdesign'] 
Installed Middleware: 
('django.middleware.common.CommonMiddleware', 
'django.contrib.sessions.middleware.SessionMiddleware', 
'django.middleware.csrf.CsrfViewMiddleware', 
'django.contrib.auth.middleware.AuthenticationMiddleware', 
'django.contrib.messages.middleware.MessageMiddleware') 


Traceback: 
File "C:\Python25\Lib\site-packages\django\core\handlers\base.py" in get_response 
    100.      response = callback(request, *callback_args, **callback_kwargs) 
File "C:\django\pley\..\pley\business\views.py" in business_view 
    57.  for place, (lat, lng) in g.geocode(string_location,exactly_one=False): 
File "c:\python25\lib\site-packages\geopy-0.94-py2.5.egg\geopy\geocoders\google.py" in geocode 
    66.   return self.geocode_url(url, exactly_one) 
File "c:\python25\lib\site-packages\geopy-0.94-py2.5.egg\geopy\geocoders\google.py" in geocode_url 
    73.   return dispatch(page, exactly_one) 
File "c:\python25\lib\site-packages\geopy-0.94-py2.5.egg\geopy\geocoders\google.py" in parse_json 
    123.   json = simplejson.loads(page) 
File "C:\Python25\lib\site-packages\simplejson-2.1.1-py2.5-win32.egg\simplejson\__init__.py" in loads 
    384.   return _default_decoder.decode(s) 
File "C:\Python25\lib\site-packages\simplejson-2.1.1-py2.5-win32.egg\simplejson\decoder.py" in decode 
    402.   obj, end = self.raw_decode(s, idx=_w(s, 0).end()) 
File "C:\Python25\lib\site-packages\simplejson-2.1.1-py2.5-win32.egg\simplejson\decoder.py" in raw_decode 
    420.    raise JSONDecodeError("No JSON object could be decoded", s, idx) 

Exception Type: JSONDecodeError at /business/view/13/ 
Exception Value: No JSON object could be decoded: line 1 column 0 (char 0) 

任何意见/答复将不胜感激。

谢谢!

+1

嗯,这是一个错误:http://code.google.com/p/geopy/issues/detail?id=15 发生变化:资源= '地图' 资源='地图/地理的作品 – wenbert 2010-07-27 15:04:21

回答

0

您是否阅读了geopy页面上的评论?

评论由gregor.horvath,09月11日,2009年

要知道,虽然有不同的 OUTPUT_FORMAT参数 的地理编码,地理编码功能的实际输出 格式 总是相同:

位置,(latttude,经度)

的OUTPUT_FORMAT仅指 地理编码服务 (Google等)的输出,而不是 地理编码功能的输出。