[英]Many to many relationship with a composite key on SQLAlchemy
假設我有以下模型:
class Molecule(Base):
db = Column(Integer, primary_key=True)
id = Column(Integer, primary_key=True)
data = Column(Integer)
class Atom(Base):
id = Column(Integer, primary_key=True)
weight = Column(Integer)
我想在 Molecule 和 Atom 之間建立多對多關系,最好的方法是什么? 注意 Molecule 的主鍵是復合的。
謝謝
多對多關聯表應該這樣定義:
molecule2atom = Table(
'molecule2atom',
Base.metadata,
Column('molecule_db', Integer),
Column('molecule_id', Integer),
Column('atom_id', Integer, ForeignKey('atom.id')),
ForeignKeyConstraint(
('molecule_db', 'molecule_id'),
('molecule.db', 'molecule.id') ),
)
並像往常一樣將關系添加到模型之一,例如,在 Class Atom 添加:
molecules = relationship("Molecule", secondary=molecule2atom, backref="atoms")
我更喜歡這里給出的解決方案 - 多對多組合鍵
如果您使用關聯表或完全聲明的表元數據,則可以在兩列中使用primary_key=True
,如建議here 。
關聯表示例:
employee_role = db.Table(
"employee_role",
db.Column("role_id", db.Integer, db.ForeignKey("role.id"), primary_key=True),
db.Column("employee_id", db.Integer, db.ForeignKey("agent.id"), primary_key=True),
)
元數據示例:
# this is using SQLAlchemy
class EmployeeRole(Base):
__tablename__ = "employee_role"
role_id = Column(Integer, primary_key=True)
employee_id = Column(Integer, primary_key=True)
# this is using Flask-SQLAlchemy with factory pattern, db gives you access to all SQLAlchemy stuff
class EmployeeRole(db.Model):
__tablename__ = "employee_role"
role_id = db.Column(db.Integer, primary_key=True)
employee_id = db.Column(db.Integer, primary_key=True)
它的 Alembic 遷移:
op.create_table(
'employee_role',
sa.Column('role_id', sa.Integer(), nullable=False),
sa.Column('employee_id', sa.Integer(), nullable=False),
sa.PrimaryKeyConstraint('role_id', 'employee_id')
)
查詢語句:
CREATE TABLE agent_role (
role_id INTEGER NOT NULL,
employee_id INTEGER NOT NULL,
PRIMARY KEY (role_id, employee_id)
);
在關系方面,在一側聲明它(這應該給你role.employees
或employee.roles
,它應該返回一個list
):
# this is using Flask-SQLAlchemy with factory pattern, db gives you access to all SQLAlchemy stuff
class Employee(db.Model):
id = db.Column(db.Integer, primary_key=True, autoincrement=True)
roles = db.relationship("Role", secondary=employee_role, backref="employee")
您的角色類可以是:
# this is using Flask-SQLAlchemy with factory pattern, db gives you access to all SQLAlchemy stuff
class Role(db.Model):
__tablename__ = "role"
id = db.Column(db.Integer, primary_key=True, autoincrement=True)
name = db.Column(db.String(25), nullable=False, unique=True)
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