Django-权限信息中间件操作

# 在当前app下新建一个middleware的文件夹,然后就可以尽情的写中间件了,只能是这个名字,切记~@!
import
re from django.shortcuts import redirect,HttpResponse from django.conf import settings class MiddlewareMixin(object): def __init__(self, get_response=None): self.get_response = get_response super(MiddlewareMixin, self).__init__() def __call__(self, request): response = None if hasattr(self, 'process_request'): response = self.process_request(request) if not response: response = self.get_response(request) if hasattr(self, 'process_response'): response = self.process_response(request, response) return response class LoginMiddleware(MiddlewareMixin): def process_request(self,request): if request.path_info == '/login/' or request.path_info == "/get_validCode_img/": return None if request.session.get('user_info'): return None return redirect('/login/') class RbacMiddleware(MiddlewareMixin): def process_request(self,request): # 1. 获取当前请求的URL # request.path_info # 2. 获取Session中保存当前用户的权限 # request.session.get("permission_url_list') current_url = request.path_info # 当前请求不需要执行权限验证 for url in settings.VALID_URL: if re.match(url,current_url): return None permission_dict = request.session.get(settings.PERMISSION_URL_DICT_KEY) if not permission_dict: return redirect('/login/') flag = False for group_id,code_url in permission_dict.items(): for db_url in code_url['urls']: regax = "^{0}$".format(db_url) if re.match(regax, current_url): request.permission_code_list = code_url['codes'] flag = True break if flag: break if not flag: return HttpResponse('无权访问')

 

posted @ 2018-03-29 14:47  前方、有光  阅读(271)  评论(0编辑  收藏  举报