[英]attribute error- current_user is NoneType in persmissions decorator
当我尝试运行我的应用程序时,我不断收到以下错误消息。
if not current_user.can(permission):
AttributeError: 'NoneType' object has no attribute 'can'
模型.py
from datetime import datetime
import hashlib
from werkzeug.security import generate_password_hash, check_password_hash
from itsdangerous import TimedJSONWebSignatureSerializer as Serializer
from markdown import markdown
import bleach
from flask import current_app, request, url_for
from flask_login import UserMixin, AnonymousUserMixin
from app.exceptions import ValidationError
from . import db, login_manager
class User(UserMixin, db.Model):
__tablename__ = 'users'
id = db.Column(db.Integer, primary_key=True)
email = db.Column(db.String(64), unique=True, index=True)
username = db.Column(db.String(64), unique=True, index=True)
role_id = db.Column(db.Integer, db.ForeignKey('roles.id'))
password_hash = db.Column(db.String(128))
confirmed = db.Column(db.Boolean, default=False)
name = db.Column(db.String(64))
location = db.Column(db.String(64))
about_me = db.Column(db.Text())
member_since = db.Column(db.DateTime(), default=datetime.utcnow)
last_seen = db.Column(db.DateTime(), default=datetime.utcnow)
avatar_hash = db.Column(db.String(32))
posts = db.relationship('Post', backref='author', lazy='dynamic')
followed = db.relationship('Follow',
foreign_keys=[Follow.follower_id],
backref=db.backref('follower', lazy='joined'),
lazy='dynamic',
cascade='all, delete-orphan')
followers = db.relationship('Follow',
foreign_keys=[Follow.followed_id],
backref=db.backref('followed', lazy='joined'),
lazy='dynamic',
cascade='all, delete-orphan')
comments = db.relationship('Comment', backref='author', lazy='dynamic')
def can(self, perm):
return self.role is not None and self.role.has_permission(perm)
def is_administrator(self):
return self.can(Permission.ADMIN)
class AnonymousUser(AnonymousUserMixin):
def can(self, permissions):
return False
def is_administrator(self):
return False
login_manager.anonymous_user = AnonymousUser
@login_manager.user_loader
def load_user(user_id):
return User.query.get(int(user_id))
装饰器.py
...
from functools import wraps
from flask import abort
from flask_login import current_user
from .models import Permission
def permission_required(permission):
def decorator(f):
@wraps(f)
def decorated_function(*args, **kwargs):
if not current_user.can(permission):
abort(403)
return f(*args, **kwargs)
return decorated_function
return decorator
def admin_required(f):
return permission_required(Permission.ADMIN)(f)
...
我的观点
@main.route('/admin_hub')
@login_required
@admin_required
def admin_hub():
return render_template('admin_hub.html')
当我尝试运行代码时,它会抛出属性错误,我不知道为什么,就好像 current_user 不存在,因为还没有用户登录,装饰器在定义时执行,导致这个错误和我真的不知道如何处理这个问题,如果你可以的话,请帮忙。 如果可以,请提供建议。 如果没有权限装饰器,代码可以完美执行。
未经测试,但是如何从flask-login.utils
导入current_user
。 这是真正的current_user
的代理,由 flask-login 在其装饰器中使用,例如login_required
。
就像是:
# Don't use current_user directly from flask_login
# from flask_login import current_user
from flask_login.utils import current_user
def permission_required(permission):
def decorator(f):
@wraps(f)
def decorated_function(*args, **kwargs):
if not current_user.can(permission):
abort(403)
return f(*args, **kwargs)
return decorated_function
return decorator
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