Question

在我看来，为什么这样做：

results = []
results.append({'status':1})
results.append({'bookmarks':[]})
simplejson.dumps(results)
# produces: []

而这不是：

from myapp.models import Bookmark
results = []
results.append({'status':1})
results.append({'bookmarks':Bookmark.objects.all()})
# fails with exception saying: [] is not JSON serializable

完全堆栈跟踪

Traceback:
File "/Users/Ishaq/Projects/github/bookmarks/venv/lib/python2.7/site-packages/django/core/handlers/base.py" in get_response
  115.                         response = callback(request, *callback_args, **callback_kwargs)
File "/Users/Ishaq/Projects/github/bookmarks/bookmarks/views.py" in index
  9.   return HttpResponse(simplejson.dumps(Bookmark.objects.all()), mimetype='application/json');
File "/usr/local/Cellar/python/2.7.3/Frameworks/Python.framework/Versions/2.7/lib/python2.7/json/__init__.py" in dumps
  231.         return _default_encoder.encode(obj)
File "/usr/local/Cellar/python/2.7.3/Frameworks/Python.framework/Versions/2.7/lib/python2.7/json/encoder.py" in encode
  201.         chunks = self.iterencode(o, _one_shot=True)
File "/usr/local/Cellar/python/2.7.3/Frameworks/Python.framework/Versions/2.7/lib/python2.7/json/encoder.py" in iterencode
  264.         return _iterencode(o, 0)
File "/usr/local/Cellar/python/2.7.3/Frameworks/Python.framework/Versions/2.7/lib/python2.7/json/encoder.py" in default
  178.         raise TypeError(repr(o) + " is not JSON serializable")

Exception Type: TypeError at /conferences/
Exception Value: [] is not JSON serializable

Answer 1

不使用simplejson来序列化django对象，而是使用s erialization provided by django.

使用参考表格链接，您可以：

from django.core import serializers
data = serializers.serialize("json", Bookmark.objects.all())

Answer 2

我在等待Burhan Khalid将他的评论转变为答案，但由于他没有，我会这样做。

使用simplejson.dumps(list(Bookmark.objects.all()))使其正常工作

django simplejson []不是JSON可序列化的

2 个答案: