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我已按照http://code.google.com/p/geopy/wiki/GettingStarted中的教程进行操作, 效果很好:

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

我想使用 json 作为输出格式,因为我想在 javascript 中处理结果。

但每次我使用:

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

我收到以下错误:

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)

任何建议/回复将不胜感激。

谢谢!

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1 回答 1

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您是否阅读了 geopy 页面上的评论?

gregor.horvath 的评论,2009 年 9 月 11 日

请注意,尽管地理编码器的 output_format 参数不同,但地理编码函数的实际输出格式始终相同:

位置,(纬度,经度)

output_format 仅指地理编码器服务(Google 等)的输出,而不是地理编码功能的输出。

于 2010-07-27T19:46:46.910 回答