6

我想在User模型中有字段，用户通过它登录username而不是email

我定义：
app.config['SECURITY_USER_IDENTITY_ATTRIBUTES'] = 'username'

但我仍然得到：

user_datastore.add_role_to_user(name, 'mgmt')
      File "/Users/boazin/sentinal/sentinel-cloud/.env/lib/python2.7/site-packages/flask_security/datastore.py", line 105, in add_role_to_user
        user, role = self._prepare_role_modify_args(user, role)
      File "/Users/boazin/sentinal/sentinel-cloud/.env/lib/python2.7/site-packages/flask_security/datastore.py", line 72, in _prepare_role_modify_args
        user = self.find_user(email=user)
      File "/Users/boazin/sentinal/sentinel-cloud/.env/lib/python2.7/site-packages/flask_security/datastore.py", line 203, in find_user
        return self.user_model.query.filter_by(**kwargs).first()
      File "/Users/boazin/sentinal/sentinel-cloud/.env/lib/python2.7/site-packages/sqlalchemy/orm/query.py", line 1333, in filter_by
        for key, value in kwargs.items()]
      File "/Users/boazin/sentinal/sentinel-cloud/.env/lib/python2.7/site-packages/sqlalchemy/orm/base.py", line 383, in _entity_descriptor
        (description, key)
    InvalidRequestError: Entity '<class 'flask_app.models.User'>' has no property 'email'

似乎电子邮件被硬编码到烧瓶安全中......

我可以改变它吗？

编辑：用户模型（根据评论中的要求）：

class User(db.Model, UserMixin):
    id = db.Column(db.Integer, primary_key=True)
    username = db.Column(db.String(255), unique=True, index=True)
    password = db.Column(db.String(255))
    token = db.Column(db.String(255), unique=True, index=True)
    active = db.Column(db.Boolean())
    confirmed_at = db.Column(db.DateTime())
    roles = db.relationship('Role', secondary=roles_users,
                            backref=db.backref('users', lazy='dynamic'))
4

3 回答 3

16

要使用用户名而不是电子邮件地址登录（使用 Flask-Security 1.7.0 或更高版本），您可以将email字段替换为模型username中的字段User

class User(db.Model, UserMixin):
    id = db.Column(db.Integer, primary_key=True)
    username = db.Column(db.String(255), unique=True, index=True)
    password = db.Column(db.String(255))
    active = db.Column(db.Boolean())
    confirmed_at = db.Column(db.DateTime())
    roles = db.relationship('Role', secondary=roles_users,
                            backref=db.backref('users', lazy='dynamic'))

并更新app配置。

app.config['SECURITY_USER_IDENTITY_ATTRIBUTES'] = 'username'

接下来，为了允许用户使用用户名而不是电子邮件登录，我们将使用LoginForm 验证方法假定用户身份属性位于email表单字段中的事实。

from flask_security.forms import LoginForm
from wtforms import StringField
from wtforms.validators import InputRequired

class ExtendedLoginForm(LoginForm):
    email = StringField('Username', [InputRequired()])

# Setup Flask-Security
user_datastore = SQLAlchemyUserDatastore(db, User, Role)
security = Security(app, user_datastore,
                    login_form=ExtendedLoginForm)

这样，我们可以使用用户名登录，而无需重写验证方法或登录模板。当然，这是一种 hack，更正确的方法是在类中添加一个检查表单字段的自定义validate方法，并相应地更新登录模板。usernameExtendedLoginForm

但是，上面的方法可以很容易地使用用户名或电子邮件地址登录。为此，请定义一个包含用户名和电子邮件字段的用户模型。

class User(db.Model, UserMixin):
    id = db.Column(db.Integer, primary_key=True)
    email = db.Column(db.String(255), unique=True)
    username = db.Column(db.String(255), unique=True, index=True)
    password = db.Column(db.String(255))
    active = db.Column(db.Boolean())
    confirmed_at = db.Column(db.DateTime())
    roles = db.relationship('Role', secondary=roles_users,
                            backref=db.backref('users', lazy='dynamic'))

并更新app配置。

app.config['SECURITY_USER_IDENTITY_ATTRIBUTES'] = ('username','email')

最后，创建自定义登录表单。

from flask_security.forms import LoginForm
from wtforms import StringField
from wtforms.validators import InputRequired

class ExtendedLoginForm(LoginForm):
    email = StringField('Username or Email Address', [InputRequired()])

# Setup Flask-Security
user_datastore = SQLAlchemyUserDatastore(db, User, Role)
security = Security(app, user_datastore,
                    login_form=ExtendedLoginForm)

现在，登录时，Flask-Security 将在电子邮件表单字段中接受电子邮件或用户名。

于 2016-04-29T04:02:44.560 回答
2

我设法通过覆盖登录表单来使用用户名或密码实现登录：

class ExtendedLoginForm(LoginForm):
    email = StringField('Username or Email Address')
    username = StringField("Username")

    def validate(self):
    from flask_security.utils import (
        _datastore,
        get_message,
        hash_password,
    )
    from flask_security.confirmable import requires_confirmation
    if not super(LoginForm, self).validate():
        return False

    # try login using email
    self.user = _datastore.get_user(self.email.data)

    if self.user is None:
        self.user = _datastore.get_user(self.username.data)

    if self.user is None:
        self.email.errors.append(get_message("USER_DOES_NOT_EXIST")[0])
        # Reduce timing variation between existing and non-existing users
        hash_password(self.password.data)
        return False
    if not self.user.password:
        self.password.errors.append(get_message("PASSWORD_NOT_SET")[0])
        # Reduce timing variation between existing and non-existing users
        hash_password(self.password.data)
        return False
    if not self.user.verify_and_update_password(self.password.data):
        self.password.errors.append(get_message("INVALID_PASSWORD")[0])
        return False
    if requires_confirmation(self.user):
        self.email.errors.append(get_message("CONFIRMATION_REQUIRED")[0])
        return False
    if not self.user.is_active:
        self.email.errors.append(get_message("DISABLED_ACCOUNT")[0])
        return False
    return True

并按照其他帖子中的说明进行注册：

# Setup Flask-Security
app.config['SECURITY_USER_IDENTITY_ATTRIBUTES'] = ('username','email')
user_datastore = SQLAlchemyUserDatastore(db, User, Role)
security = Security(app, user_datastore,
                login_form=ExtendedLoginForm)

由于电子邮件和用户名是可选的，现在其中之一可用于登录。但请确保这两个字段在 DB 模型中都设置为唯一的。

于 2020-08-15T10:30:02.303 回答
1

来自https://pythonhosted.org/Flask-Security/models.html

字段id, email, password, active是必不可少的。所以添加

email = db.Column(db.String(255), unique=True)

只需username沿此添加您的自定义字段。

于 2015-06-14T11:27:41.043 回答