Question

我有一个Django应用程序。在视图中，我调用另一个函数（在stats.py中）然后进行HTTP POST。

views.py

  from stats import Stat
  a = Stat(example="12345")
  a.use(id='query')

stats.py

   self.data = { example : "12345" }

   req = urllib2.Request(api_url)
   req.add_header('Content-Type', 'application/json')
   response = urllib2.urlopen(req, json.dumps(self.data))

出现的问题是我收到错误

<django.utils.functional.SimpleLazyObject object at 0x2b4d1fe47650> is not JSON serializable

Django Traceback

从查看Django Traceback，我得到以下内容，

 /prod/tools/lx/views.py in update_input

            a.use(id='query')

     ...

/prod/tools/main/stats.py in log_use

            response = urllib2.urlopen(req, json.dumps(self.data))

     ...

/usr/local/lib/python2.7/json/__init__.py in dumps

            return _default_encoder.encode(obj)

     ...

/usr/local/lib/python2.7/json/encoder.py in encode

            chunks = self.iterencode(o, _one_shot=True)

     ...

/usr/local/lib/python2.7/json/encoder.py in iterencode

            return _iterencode(o, 0)

     ...

/usr/local/lib/python2.7/json/encoder.py in default

            raise TypeError(repr(o) + " is not JSON serializable")

     ...

有什么想法吗？

谢谢，

Answer 1

使用urllib

尝试此操作

import urllib
...
    self.data = { example : "12345", 'Content-type':'application/json' }    
    self.data = urllib.urlencode(self.data)
    req = urllib2.Request(api_url, self.data)
    response = urllib2.urlopen(req)

<django.utils.functional.simplelazyobject object =“”at =“”0x2b4d1fe47650 =“”>不是JSON可序列化的

1 个答案: