Python中的simplejson抛出值错误

时间:2011-09-22 09:58:48

标签: python json google-app-engine decode

我有一个JSON字符串,我将其发布到我的Python脚本中。这是一个示例字符串:

{"uid":"1111111","method":"check_user"}

在我的Python代码中,我只需调用simplejson.loads( str ),其中str是请求中的JSON字符串。 JSON字符串看起来很好,因为当我在请求时打印它时它完好无损。但是我得到了一个ValueError:

Extra data: line 1 column 41 - line 1 column 48 (char 41 - 48)
Traceback (most recent call last):   File
"/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/ext/webapp/_webapp25.py",
line 703, in __call__
    handler.post(*groups)   File "/Users/.../app/controller/api_controller.py", line 25, in post
    req = simplejson.loads( req )   File
"/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/lib/django_0_96/django/utils/simplejson/__init__.py",
line 232, in loads
    return cls(encoding=encoding, **kw).decode(s)   File
"/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/lib/django_0_96/django/utils/simplejson/decoder.py",
line 254, in decode
    raise ValueError(errmsg("Extra data", s, end, len(s)))

任何想法可能是什么?我尝试从字符串中删除新的行,制表符和斜杠,甚至使用.decode('string_escape')

对其进行解码

1 个答案:

答案 0 :(得分:6)

你的字符串中有一些不可打印的字符。如果我将一个空字节附加到字符串的末尾,并且print它没有显示问题,我会得到相同的错误:

>>> import json
>>> string = '{"uid":"1111111","method":"check_user"}\x00'
>>> print string
{"uid":"1111111","method":"check_user"}
>>> print repr(string)
'{"uid":"1111111","method":"check_user"}\x00'
>>> json.loads(string)
Traceback (most recent call last):
  File "<interactive input>", line 1, in <module>
  File "C:\Python27\Lib\json\__init__.py", line 326, in loads
    return _default_decoder.decode(s)
  File "C:\Python27\Lib\json\decoder.py", line 369, in decode
    raise ValueError(errmsg("Extra data", s, end, len(s)))
ValueError: Extra data: line 1 column 39 - line 1 column 40 (char 39 - 40)

在请求时打印字符串的repr,你应该看到它。