Question

我有一个与2个表相关的数据透视表（postgres）。我想使用flask-marshmallow和sqlalchemy从两个表中的任何棉花糖架构中获取数据。例如：Table1Schema（）。dump（table1_object）.first：并获得与Table2的数据内部连接的Table1记录（Many = True）：下面是我当前的代码：

仅供参考：我对烧瓶和ORM世界是陌生的：

我的模特：

class Permission(db.Model):
    __tablename__ = 'permission'
    permission_id = db.Column(db.Integer, primary_key=True)
    object = db.Column(db.String(70), nullable=False)

    def __init__(self, object):
        self.object = object


class Role(db.Model):
    __tablename__ = 'role'
    role_id = db.Column(db.Integer, primary_key = True)
    role_name = db.Column(db.String(70), nullable=False)
    role_permissions = db.relationship("RolePermission",backref='Role', lazy='dynamic')

    def __init__(self,role_name):
        self.role_name = role_name



class RolePermission(db.Model):
    __tablename__ = 'role_permission'
    role_permission_id = db.Column(db.Integer, primary_key=True)
    role_id = db.Column(db.Integer, db.ForeignKey('role.role_id', ondelete='CASCADE'), nullable =False)
    permission_id = db.Column(db.Integer, db.ForeignKey('permission.permission_id', ondelete='CASCADE'), nullable =False)

    def __init__(self,role_id,permission_id):
        self.role_id = role_id
        self.permission_id = permission_id

我的模式：

class RoleSchema(ma.Schema):
    role_id = fields.Integer(dump_only=True)
    role_name =  fields.String(required=True, validate=validate.Length(1))
    permissions = fields.Nested('RolePermissionSchema', many=True, only=('permission',))

    class Meta:
        model = Role
        fields = ('role_id', 'role_name', 'permissions')


class RolePermissionSchema(ma.Schema):
    role_permission_id = fields.Integer(dump_only=True)
    role_id = fields.Integer(required=True)
    permission_id = fields.Integer(required=True)
    role = fields.Nested('ROleSchema', many=False, only=('role_id', 'role_name',))
    permission = fields.Nested('PermissionSchema', many=False, only=('object', 'action',))




class PermissionSchema(ma.Schema):
    permission_id = fields.Integer(dump_only=True)
    object = fields.String(required=True, validate=validate.Length(1))
    role_permissions = fields.Nested('RolePermissionSchema', many=True, only=('role_permission_id', 'role_id',))
    class Meta:
        fields = ('object','action','role_permissions',)

序列化器：

 role = Role.query.filter_by(role_name=data['role_name']).filter_by(status=data['status']).first()
role_schema.dump(role).data

我要打印一个具有opermission []对象的Role o对象。但是，从以上我只能得到Role内容。下面是输出：

{
        "role_name": "user",
        "status": true,
        "role_id": 4
    }

我如何得到这样的东西：

{
    "role_name": "user",
    "status": true,
    "role_id": 4,
    "permissions":[{
         "object":"user",
         "action":"create"},
          ]
}

Answer 1

有一些命名的问题，通过注释行这里表示#。

class Role(db.Model):
   __tablename__ = 'role'
   role_id = db.Column(db.Integer, primary_key = True)
   role_name = db.Column(db.String(70), nullable=False)
   # This backref should probably be named role
   role_permissions = db.relationship("RolePermission",backref='Role', lazy='dynamic')

class RoleSchema(ma.Schema):
    role_id = fields.Integer(dump_only=True)
    role_name =  fields.String(required=True, validate=validate.Length(1))
    # This does not correspond to role_permissions in class Role
    permissions = fields.Nested('RolePermissionSchema', many=True, only=('permission',))

    class Meta:
        model = Role
        fields = ('role_id', 'role_name', 'permissions')

class RolePermissionSchema(ma.Schema):
   role_permission_id = fields.Integer(dump_only=True)
   role_id = fields.Integer(required=True)
   permission_id = fields.Integer(required=True)
   # Should be RoleSchema
   role = fields.Nested('ROleSchema', many=False, only=('role_id', 'role_name',))
   permission = fields.Nested('PermissionSchema', many=False, only=('object', 'action',))

嵌套棉花糖和Sqlalchemy包含子项

1 个答案: