如何在 django 函数中将单个对象作为 json 响应返回？

Question

我只想将 User 模型的单个对象作为 json 响应。我不知道我该怎么做..

现在我收到“用户”对象不可迭代的错误

这是我的功能：

def some_view(username_to_toggle):
    print(username_to_toggle,"User to togle")
    user = User.objects.get(username__iexact=username_to_toggle)
    print(user,"User object")
    user_json = serializers.serialize('json', user,many=False)
    print(user,"User json")
    return HttpResponse(user_json, content_type='application/json')

追溯：

Traceback (most recent call last):
  File "/home/hamza/Mydjangoproject/grabpublic/grabpublic/Grab_env/lib/python3.6/site-packages/django/core/handlers/exception.py", line 34, in inner
    response = get_response(request)
  File "/home/hamza/Mydjangoproject/grabpublic/grabpublic/Grab_env/lib/python3.6/site-packages/django/core/handlers/base.py", line 115, in _get_response
    response = self.process_exception_by_middleware(e, request)
  File "/home/hamza/Mydjangoproject/grabpublic/grabpublic/Grab_env/lib/python3.6/site-packages/django/core/handlers/base.py", line 113, in _get_response
    response = wrapped_callback(request, *callback_args, **callback_kwargs)
  File "/home/hamza/Mydjangoproject/grabpublic/grabpublic/Grab_env/lib/python3.6/site-packages/django/views/generic/base.py", line 71, in view
    return self.dispatch(request, *args, **kwargs)
  File "/home/hamza/Mydjangoproject/grabpublic/grabpublic/Grab_env/lib/python3.6/site-packages/django/contrib/auth/mixins.py", line 52, in dispatch
    return super().dispatch(request, *args, **kwargs)
  File "/home/hamza/Mydjangoproject/grabpublic/grabpublic/Grab_env/lib/python3.6/site-packages/django/views/generic/base.py", line 97, in dispatch
    return handler(request, *args, **kwargs)
  File "/home/hamza/Mydjangoproject/grabpublic/grabpublic/profiles/views.py", line 65, in post
    profile_, is_following,json_follower = UserProfile.objects.toggle_follow(request.user, request.user.id ,username_to_toggle,json_follower)
  File "/home/hamza/Mydjangoproject/grabpublic/grabpublic/profiles/models.py", line 72, in toggle_follow
    json_follower = some_view(username_to_toggle)
  File "/home/hamza/Mydjangoproject/grabpublic/grabpublic/profiles/models.py", line 45, in some_view
    user_json = serializers.serialize('json', user,many=False)
  File "/home/hamza/Mydjangoproject/grabpublic/grabpublic/Grab_env/lib/python3.6/site-packages/django/core/serializers/__init__.py", line 128, in serialize
    s.serialize(queryset, **options)
  File "/home/hamza/Mydjangoproject/grabpublic/grabpublic/Grab_env/lib/python3.6/site-packages/django/core/serializers/base.py", line 90, in serialize
    for count, obj in enumerate(queryset, start=1):
TypeError: 'User' object is not iterable

Answer 1

来自django.core.serializers 的serialize 方法没有many 选项，它总是需要一个模型对象的列表或查询集来进行序列化。如果您总是只想序列化一个对象，请尝试使用其中任何一个：

第一种方法，使用查询集：

def some_view(username_to_toggle):
    print(username_to_toggle,"User to togle")
    users = User.objects.filter(username__iexact=username_to_toggle)
    print(users,"User objects")
    user_json = serializers.serialize('json', users)
    print(user,"User json")
    return HttpResponse(user_json, content_type='application/json')

或第二种方法，使用列表：

def some_view(username_to_toggle):
    print(username_to_toggle,"User to togle")
    user = User.objects.get(username__iexact=username_to_toggle)
    print(user,"User object")
    user_json = serializers.serialize('json', [user])
    print(user,"User json")
    return HttpResponse(user_json, content_type='application/json')

many=False 或 many=True 是 Django REST Framework 序列化器的一个特性，但它们的用法不同且更复杂，因为您需要为要序列化的每个模型声明一个单独的序列化器。