Question

我有一个用烧瓶建造的应用民宿。每个寄宿家庭都有一个用户登录名，该登录名通过flask-security构建，每个寄宿家庭所有者均具有角色用户。每个用户都可以使用flask-admin输入自己的寄宿家庭数据。但不幸的是，如果一个用户输入了他们的数据，则其他具有角色 User 的用户也可以输入数据。

所以..我的问题是，如果用户具有相同的角色，该如何分隔数据？用户A只能看到他的数据，用户B也可以看到。.

这是我的 models.py 代码：

    roles_users = database.Table(
    'roles_users',
    database.Column('user_id', database.Integer(), database.ForeignKey('user.id')),
    database.Column('role_id', database.Integer(), database.ForeignKey('role.id'))
)


class Role(database.Model, RoleMixin):
    id = database.Column(database.Integer(), primary_key=True)
    name = database.Column(database.String(80), unique=True)
    description = database.Column(database.String(255))

    def __str__(self):
        return self.name


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

    def __str__(self):
        return self.email

class Room(database.Model):
    __tablename__ = 'room'
    room_id = Column(Integer, primary_key=True)
    room_name = Column(String)
    room_description = Column(String)
    room_images = Column(database.Unicode(128))
    room_price = Column(Integer) 
    user_id = Column(Integer, ForeignKey(User.id))

这是我的 views.py ：

class UserAccess(ModelView):

    def is_accessible(self):
        if not current_user.is_active or not current_user.is_authenticated:
            return False
        if current_user.has_role('user'):
            return True

        return False

    def _handle_view(self, name, **kwargs):
        """
        Override builtin _handle_view in order to redirect users when a view is not accessible.
        """
        if not self.is_accessible():
            if current_user.is_authenticated:
                # permission denied
                abort(403)
            else:
                # login
                return redirect(url_for('security.login', next=request.url))

class Room(UserAccess):
    form_overrides = dict(keterangan_kamar=CKEditorField)
    create_template = 'admin/ckeditor.html'
    edit_template = 'admin/ckeditor.html'
    column_list = ('room_name', 'room_images', 'room_price')
    def _list_thumbnail(view, context, model, name):
        if not model.room_images:
            return ''

        return Markup('<img src="%s">' % url_for('static',
                                                 filename=form.thumbgen_filename(model.room_images)))

    column_formatters = {
        'room_images': _list_thumbnail
    }

    # Alternative way to contribute field is to override it completely.
    # In this case, Flask-Admin won't attempt to merge various parameters for the field.
    form_extra_fields = {
        'room_images': form.ImageUploadField('Room Images',
                                      base_path=file_path,
                                      thumbnail_size=(100, 100, True))
    }

Answer 1

您需要通过覆盖get_query和get_count_query

向您的Room视图添加永久过滤器

例如（请注意，我已经将您的Room类命名为RoomView，因为您已经有一个名为Room的代表数据库模型的类）：

class RoomView(UserAccess):

    def get_query(self):
        return self.session.query(self.model).filter(
            Room.user_id == current_user.id
        )

    def get_count_query(self):
        return self.session.query(func.count('*')).select_from(self.model).filter(
            Room.user_id == current_user.id
        )

    #  ... your code etc

烧瓶安全性分隔烧瓶管理员上的访问数据

1 个答案: