django - django 中 pre_delete 信号中的当前用户

Question

是否可以在连接到 pre_delete 信号的回调中获取登录的 django 用户(调用模型的删除方法)？

Answer 1

pre_delete 信号不会传递请求实例，但您可以添加一个装饰器来添加它，并在 View 上使用该装饰器，这会删除指定的模型。

假设这是回调函数:

def pre_delete_cb(sender, instance, using, **kwargs):
    pass

正在装饰器中添加:

from django.db.models.signals import pre_delete
from functools import wraps
from django.utils.decorators import available_attrs

def pre_delete_dec(cb, sender):
    def decorator(view_func):
        @wraps(view_func, assigned=available_attrs(view_func))
        def _wrapped_view(request, *args, **kwargs):
            cb.request = request  # here we add the request to the callback function
            pre_delete.connect(cb, sender=sender)  # and connecting the real pre_delete to the callback
            return view_func(request, *args, **kwargs)
        return _wrapped_view
    return decorator

并以您调用的方式在 View 上使用装饰器 - 而不是 pre_delete.connect(pre_delete_cb, MyModel)，使用:

@pre_delete_dec(pre_delete_cb, MyModel)
def myview(request):

然后在回调中您可以通过以下方式访问请求:

def pre_delete_cb(sender, instance, using, **kwargs):
    current_user = pre_delete_cb.request.user

您可以在全局级别添加此内容，而不仅仅是每个 View - 使用中间件:

from django.db.models.signals import pre_delete

def pre_delete_cb(sender, instance, using, **kwargs):
    current_user = pre_delete_cb.request.user

class PreDeleteMiddleware(object):
    def process_view(self, request, view_func, view_args, view_kwargs):
        pre_delete_cb.request = request
        from myapp.models import MyModel
        pre_delete.connect(pre_delete_cb, sender=MyModel)