Django 1.7引入了JsonResponse objects,我尝试使用它将值列表返回给我的ajax请求。如何将模型字段传递给JsonResponse对象
我想通过
>>> Genre.objects.values('name', 'color')
[{'color': '8a3700', 'name': 'rock'}, {'color': 'ffff00', 'name': 'pop'}, {'color': '8f8f00', 'name': 'electronic'}, {'color': '9e009e', 'name': 'chillout'}, {'color': 'ff8838', 'name': 'indie'}, {'color': '0aff0a', 'name': 'techno'}, {'color': 'c20000', 'name': "drum'n'bass"}, {'color': '0000d6', 'name': 'worldmusic'}, {'color': 'a800a8', 'name': 'classic'}, {'color': 'dbdb00', 'name': 'hiphop'}]
到JsonResponse对象。
但是,我的尝试失败。
>>> JsonResponse({'foo': 'bar', 'blib': 'blab'}) # works
<django.http.response.JsonResponse object at 0x7f53d28bbb00>
>>> JsonResponse(Genre.objects.values('name', 'color')) # doesn't work
Traceback (most recent call last):
File "<console>", line 1, in <module>
File "/home/marcel/Dokumente/django/FlushFM/env/lib/python3.4/site-packages/django/http/response.py", line 476, in __init__
raise TypeError('In order to allow non-dict objects to be '
TypeError: In order to allow non-dict objects to be serialized set the safe parameter to False
这可能是由于Genre.objects.values()
的数据结构不同所致。
这将如何做对吗?
[编辑]
随着safe=False
我得到
>>> JsonResponse(Genre.objects.values('name', 'color'), safe=False)
Traceback (most recent call last):
File "<console>", line 1, in <module>
File "/home/marcel/Dokumente/django/FlushFM/env/lib/python3.4/site-packages/django/http/response.py", line 479, in __init__
data = json.dumps(data, cls=encoder)
File "/usr/lib/python3.4/json/__init__.py", line 237, in dumps
**kw).encode(obj)
File "/usr/lib/python3.4/json/encoder.py", line 192, in encode
chunks = self.iterencode(o, _one_shot=True)
File "/usr/lib/python3.4/json/encoder.py", line 250, in iterencode
return _iterencode(o, 0)
File "/home/marcel/Dokumente/django/FlushFM/env/lib/python3.4/site-packages/django/core/serializers/json.py", line 109, in default
return super(DjangoJSONEncoder, self).default(o)
File "/usr/lib/python3.4/json/encoder.py", line 173, in default
raise TypeError(repr(o) + " is not JSON serializable")
TypeError: [{'color': '8a3700', 'name': 'rock'}, {'color': 'ffff00', 'name': 'pop'}, {'color': '8f8f00', 'name': 'electronic'}, {'color': '9e009e', 'name': 'chillout'}, {'color': 'ff8838', 'name': 'indie'}, {'color': '0aff0a', 'name': 'techno'}, {'color': 'c20000', 'name': "drum'n'bass"}, {'color': '0000d6', 'name': 'worldmusic'}, {'color': 'a800a8', 'name': 'classic'}, {'color': 'dbdb00', 'name': 'hiphop'}] is not JSON serializable
什么工作是
>>> JsonResponse(list(Genre.objects.values('name', 'color')), safe=False)
<django.http.response.JsonResponse object at 0x7f53d28bb9e8>
但是是不是有更好的办法来产生一个字典出一个模型对象的?
您是否尝试了解错误信息所说的内容? – 2014-09-26 19:59:11
@DanielRoseman很好,但是也设置'safe = False'会导致错误信息。你需要输出吗? – speendo 2014-09-26 20:02:53
@speendo与'safe = False'同样的错误呢?也许尝试将'ValuesQuerySet'传递给'list()':''JsonResponse(list(Genre.objects.values('name','color')))' – 2014-09-26 20:06:17