在AuthenticationMiddleware中添加user_id属性| Django的

问题描述

关于Django身份验证中间件，我有以下问题：

class AuthenticationMiddleware(MiddlewareMixin):
    def process_request(self,request):
        assert hasattr(request,'session'),(
            "The Django authentication middleware requires session middleware "
            "to be installed. Edit your MIDDLEWARE setting to insert "
            "'django.contrib.sessions.middleware.SessionMiddleware' before "
            "'django.contrib.auth.middleware.AuthenticationMiddleware'."
        )
        request.user = SimpleLazyObject(lambda: get_user(request))

正如您在此处看到的那样，中间件从后端调用方法get_user（在我的情况下为simple_jwt），并将此方法放入SimpleLazyObject中，以便稍后进行评估。

def get_user(self,validated_token):
    """
    Attempts to find and return a user using the given validated token.
    """
    try:
        user_id = validated_token[api_settings.USER_ID_CLaim]
    except KeyError:
        raise InvalidToken(_('Token contained no recognizable user identification'))

    try:
        user = User.objects.get(**{api_settings.USER_ID_FIELD: user_id})
    except User.DoesNotExist:
        raise AuthenticationFailed(_('User not found'),code='user_not_found')

    if not user.is_active:
        raise AuthenticationFailed(_('User is inactive'),code='user_inactive')

    return user

我要做的是保留request.user = SimpleLazyObject(lambda: get_user(request))，以便为使用它的应用程序提供用户实例，但对于我的自定义应用程序，我想添加类似的内容

伪代码

request.user_id = user_id from  user (user_id = validated_token[api_settings.USER_ID_CLaim])

为了不查询数据库中对用户对象的每一个请求，以防万一我只需要user_id，它已经直接在后端的get_user方法中提供了。

问题–如何将user_id从get_user()传递到AuthenticationMiddleware.process_request()并设置request.user_id属性以请求而不评估SimpleLazyObject？

奇怪的是，我无法在class JWTAuthentication(authentication.BaseAuthentication):所属的get_user()中分配要请求的属性

谢谢。

Internal Server Error: /auth/users/me/
Traceback (most recent call last):
  File "C:\ProgramData\Miniconda3\envs\entropy\lib\site-packages\asgiref\sync.py",line 330,in thread_handler
    raise exc_info[1]
  File "C:\ProgramData\Miniconda3\envs\entropy\lib\site-packages\django\core\handlers\exception.py",line 38,in inner
    response = await get_response(request)
  File "C:\ProgramData\Miniconda3\envs\entropy\lib\site-packages\django\utils\deprecation.py",line 126,in __acall__
    response = await sync_to_async(
  File "C:\ProgramData\Miniconda3\envs\entropy\lib\site-packages\asgiref\sync.py",line 296,in __call__
    ret = await asyncio.wait_for(future,timeout=None)
  File "C:\ProgramData\Miniconda3\envs\entropy\lib\asyncio\tasks.py",line 440,in wait_for
    return await fut
  File "C:\ProgramData\Miniconda3\envs\entropy\lib\site-packages\asgiref\current_thread_executor.py",line 23,in run
    result = self.fn(*self.args,**self.kwargs)
  File "C:\ProgramData\Miniconda3\envs\entropy\lib\site-packages\asgiref\sync.py",line 334,in thread_handler
    return func(*args,**kwargs)
  File "C:\ProgramData\Miniconda3\envs\entropy\lib\site-packages\django\contrib\auth\middleware.py",line 26,in process_request
    request.user_id = _get_user_session_key(request)
  File "C:\ProgramData\Miniconda3\envs\entropy\lib\site-packages\django\contrib\auth\__init__.py",line 58,in _get_user_session_key
    return get_user_model()._Meta.pk.to_python(request.session[SESSION_KEY])
  File "C:\ProgramData\Miniconda3\envs\entropy\lib\site-packages\django\contrib\sessions\backends\base.py",line 65,in __getitem__
    return self._session[key]
KeyError: '_auth_user_id'

解决方法

您已经具有可用的用户ID。它是从auth.get_user（）中的会话中解码的，您可以将其复制到自己的MiddleWare中：

from django.contrib.auth import _get_user_session_key

request.user_id = _get_user_session_key(request)

这是Django私有API，不确定为什么将其保持私有。但是您也可以复制它实现的一种衬纸：

request.user_id = get_user_model()._meta.pk.to_python(
    request.session[SESSION_KEY]
)