我有一个django应用程序,它运行在1.4.2
版本并且工作正常,但最近我将其更新为django 1.6.5
并面临一些奇怪的错误,如下所示
实际上我在我的网站功能
中的用户/客户注册过程中得到了这个Request URL: http://example.com/client/registration/
Django Version: 1.6.5
Exception Type: TypeError
Exception Value: <Client: test one> is not JSON serializable
Exception Location: /usr/lib/python2.7/json/encoder.py in default, line 184
Python Executable: /home/user/.virtualenvs/test_proj/bin/python
Python Version: 2.7.5
回溯
Traceback:
File "/home/user/.virtualenvs/test_proj/local/lib/python2.7/site-packages/django/core/handlers/base.py" in get_response
199. response = middleware_method(request, response)
File "/home/user/.virtualenvs/test_proj/local/lib/python2.7/site-packages/django/contrib/sessions/middleware.py" in process_response
38. request.session.save()
File "/home/user/.virtualenvs/test_proj/local/lib/python2.7/site-packages/django/contrib/sessions/backends/db.py" in save
57. session_data=self.encode(self._get_session(no_load=must_create)),
File "/home/user/.virtualenvs/test_proj/local/lib/python2.7/site-packages/django/contrib/sessions/backends/base.py" in encode
87. serialized = self.serializer().dumps(session_dict)
File "/home/user/.virtualenvs/test_proj/local/lib/python2.7/site-packages/django/core/signing.py" in dumps
88. return json.dumps(obj, separators=(',', ':')).encode('latin-1')
File "/usr/lib/python2.7/json/__init__.py" in dumps
250. sort_keys=sort_keys, **kw).encode(obj)
File "/usr/lib/python2.7/json/encoder.py" in encode
207. chunks = self.iterencode(o, _one_shot=True)
File "/usr/lib/python2.7/json/encoder.py" in iterencode
270. return _iterencode(o, 0)
File "/usr/lib/python2.7/json/encoder.py" in default
184. raise TypeError(repr(o) + " is not JSON serializable")
Exception Type: TypeError at /client/registration/
Exception Value: <Client: test one> is not JSON serializable
我很困惑为什么上面的json错误在更新后出现,以及我在某些模型中使用自定义json字段的方式如下所示
<强烈>凸出/ utils.py
from django.db import models
from django.utils import simplejson as json
from django.core.serializers.json import DjangoJSONEncoder
class JSONField(models.TextField):
'''JSONField is a generic textfield that neatly serializes/unserializes
JSON objects seamlessly'''
# Used so to_python() is called
__metaclass__ = models.SubfieldBase
def to_python(self, value):
'''Convert our string value to JSON after we load it from the DB'''
if value == '':
return None
try:
if isinstance(value, basestring):
return json.loads(value)
except ValueError:
pass
return value
def get_db_prep_save(self, value, connection=None):
'''Convert our JSON object to a string before we save'''
if not value or value == '':
return None
if isinstance(value, (dict, list)):
value = json.dumps(value, mimetype="application/json")
return super(JSONField, self).get_db_prep_save(value, connection=connection)
from south.modelsinspector import add_introspection_rules
add_introspection_rules([], ["^proj\.util\.jsonfield\.JSONField"])
settings.py
SERIALIZATION_MODULES = {
'custom_json': 'proj.util.json_serializer',
}
json_serializer.py
from django.core.serializers.json import Serializer as JSONSerializer
from django.utils.encoding import is_protected_type
# JSONFields that are normally incorrectly serialized as strings
json_fields = ['field_1', 'field_2']
class Serializer(JSONSerializer):
"""
A fix on JSONSerializer in order to prevent stringifying JSONField data.
"""
def handle_field(self, obj, field):
value = field._get_val_from_obj(obj)
# Protected types (i.e., primitives like None, numbers, dates,
# and Decimals) are passed through as is. All other values are
# converted to string first.
if is_protected_type(value) or field.name in json_fields:
self._current[field.name] = value
else:
self._current[field.name] = field.value_to_string(obj)
那么如何解决上述错误?有人可以给我解释导致错误的原因吗?
答案 0 :(得分:42)
Django 1.6将序列化器从pickle改为json。 pickle可以序列化json所能做的事情。
您可以更改settings.py
中SESSION_SERIALIZER的值,以便在版本1.6之前从Django恢复行为。
SESSION_SERIALIZER = 'django.contrib.sessions.serializers.PickleSerializer'
您可能希望阅读文档中的session serialization。
答案 1 :(得分:5)
将此行设置为settings.py将在升级到django 1.6版本时清除错误
SESSION_SERIALIZER = 'django.contrib.sessions.serializers.PickleSerializer'
答案 2 :(得分:1)
在分析回溯后,似乎JSONEncoder无法序列化客户端模型的实例。通常,如果您尝试使用 json 或 simplejson 库序列化与其他模型(Many2ManyField等)相关的模型,则会出现此类错误。
请参阅此https://docs.djangoproject.com/en/dev/topics/serialization/,您还可以根据需要使用某些第三方软件包,例如 DjangoRestFramework 。