[英]Query many-to-many in SQLAlchemy
我有以下型號。 用戶具有多個角色,並且一個角色可以具有許多權限。 我不太明白如何查詢以獲得想要的東西。
user_role = db.Table(
'user_role',
db.Column('user_id', db.Integer, db.ForeignKey('user.id')),
db.Column('role_id', db.Integer, db.ForeignKey('role.id')),
db.UniqueConstraint('user_id', 'role_id')
)
role_permission = db.Table(
'role_permission',
db.Column('permission_id', db.Integer, db.ForeignKey('permission.id')),
db.Column('role_id', db.Integer, db.ForeignKey('role.id')),
db.UniqueConstraint('permission_id', 'role_id')
)
class Role(Base):
__tablename__ = 'role'
id = db.Column(db.Integer, primary_key=True)
name = db.Column(db.String(100), unique=True, nullable=False)
class Permission(Base):
__tablename__ = 'permission'
id = db.Column(db.Integer, primary_key=True)
name = db.Column(db.String(100), nullable=False)
roles = db.relation(Role, secondary=role_permission, backref=db.backref('permissions'))
class User(Base, UserMixin):
__tablename__ = 'user'
id = db.Column(db.Integer, primary_key=True)
username = db.Column(db.String(60), unique=True, nullable=False)
password_hash = db.Column(db.String(80), nullable=False)
roles = db.relation(Role, secondary=user_role, backref=db.backref('users'))
我想獲得分配給用戶的所有權限的列表(最好是唯一的),但是我似乎無法弄清楚該如何做。
我可以通過在User模型中創建一個生成器來獲取列表:
def get_all_permissions(self):
for role in self.roles:
for perm in role.permissions:
yield perm
但我很希望能夠在一個查詢中做到這一點。
好吧,僅獲取權限列表,請嘗試如下操作:
permissions = session.query(Permission).\
join(Role).join(User).filter(User.username='MisterX').all()
或過濾您想要的任何東西。 要使權限唯一,可以使用group by:
permissions = session.query(Permission.id, Permission.name).join(Role).join(User).\
filter(User.username='MisterX').group_by(Permission.id).all()
或者,如果沒有特殊查詢,則可以使用聲明性擴展:
permissions = User.roles.permissions
有幫助嗎?
您未指定元數據參數,可能無法正確識別關聯表。 該腳本對我有用:
#!/bin/python
from sqlalchemy import Table
from sqlalchemy import Integer, String, ForeignKey, create_engine, Column, PrimaryKeyConstraint
from sqlalchemy.orm import relationship, backref, sessionmaker
from sqlalchemy.ext.declarative import declarative_base
engine = create_engine('sqlite:///:memory:', echo=True)
Base = declarative_base()
user_role = Table(
'user_role',
Base.metadata,
Column('user_id', Integer, ForeignKey('users.id')),
Column('role_id', Integer, ForeignKey('roles.id')),
PrimaryKeyConstraint('user_id', 'role_id')
)
role_permission = Table(
'role_permission',
Base.metadata,
Column('permission_id', Integer, ForeignKey('permissions.id')),
Column('role_id', Integer, ForeignKey('roles.id')),
PrimaryKeyConstraint('permission_id', 'role_id')
)
class Role(Base):
__tablename__ = 'roles'
id = Column(Integer, primary_key=True)
name = Column(String(100), unique=True, nullable=False)
class Permission(Base):
__tablename__ = 'permissions'
id = Column(Integer, primary_key=True)
name = Column(String(100), nullable=False)
roles = relationship("Role", secondary=role_permission, backref=backref('permissions'))
class User(Base):
__tablename__ = 'users'
id = Column(Integer, primary_key=True)
username = Column(String(60), unique=True, nullable=False)
password_hash = Column(String(80), nullable=False)
roles = relationship("Role", secondary=user_role, backref=backref('users'))
Base.metadata.create_all(engine)
session = sessionmaker(bind=engine)()
u = User(username="user", password_hash="secret")
r1 = Role(name="Role 1")
session.add(r1)
r2 = Role(name="Role 2")
session.add(r2)
p1 = Permission(name="Permission 1")
p2 = Permission(name="Permission 2")
p3 = Permission(name="Permission 3")
r1.permissions.append(p1)
r1.permissions.append(p2)
r2.permissions.append(p2)
r2.permissions.append(p3)
u.roles.append(r1)
u.roles.append(r2)
session.add(u)
for perm in session.query(Permission).join(Role, Permission.roles).\
join(User, Role.users).filter(User.username=="user").distict()all():
print(perm.name)
如果您已經將User對象以及Permissions和Roles一起加載到內存中,則您的代碼應快速完成操作,而無需進入數據庫。
否則,此查詢應工作:
user_id = 789
permissions = (db.session.query(Permission)
.join(Role, Permission.roles)
.join(User, Role.users)
.filter(User.id == user_id)
).distinct()
#print(permissions)
for perm in permissions:
print(perm)
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