Python SQLAlchemy一对一关系创建多行

我试图在我的数据库（postgresql）中的两个表之间建立一对一的关系。我正在python中使用SQLAlchemy。因此，我使用了文档中给出的一个示例

这将创建两个表parent和child。 现在我在父表和子表中插入值

from sqlalchemy import create_engine
from sqlalchemy.orm import sessionmaker
from test_databasesetup import Base, Parent, Child

engine = create_engine('postgresql+psycopg2://testdb:hello@localhost/fullstack')
Base.metadata.bind = engine
DBSession = sessionmaker(bind=engine)
session = DBSession()

parent = Parent()
session.add(parent)
session.commit() // insert in 'parent' table with id=1
// insert into child
child = Child(parent_id=1)
session.add(child)
session.commit()

child = Child(parent_id=1)
session.add(child)
session.commit() 

再次插入具有相同父\u id的子项本应引发错误，但已插入记录

id | parent_id 
----+-----------
  1 |         1
  2 |         1

这里应该做些什么，以便我只能插入一个与父id对应的子项。我不希望子项具有相同的父id

---

谢谢。

问题是您直接指定了字段

parent\u id

。在这种情况下，

sqlalchemy

没有机会验证一对一的关系。相反，请使用以下关系：

# add new parent to the database
parent = Parent()
session.add(parent)
session.commit()

# insert new child for this parent
child = Child(parent=parent)  # @note: this is the code change. \
# Here we refer to parent, not parent_id field
session.add(child)
session.commit()

# insert another child for the same parent:
# this will assign this new child to the parent, and will 
# *remove* previous child from this parent
child = Child(parent=parent)
session.add(child)
session.commit()

另一个副作用是代码更加简洁。还有一个事实是

sqlalchemy

可以识别外键本身，您不需要知道对象的

id

：

parent = Parent()
child = Child(parent=parent)
# it is enough to add only one related object, and the other (Child) will be added to session automatically
session.add(parent)
# we can commit only now, when we finished working on all the objects in the current session/transaction
session.commit()

---

此外，您可以将唯一约束添加到

Child.parent\u id

字段，作为附加的数据库级检查，这将在您的情况下引发错误：

parent = Parent()
child = Child(parent=parent)
# it is enough to add only one related object, and the other (Child) will be added to session automatically
session.add(parent)
# we can commit only now, when we finished working on all the objects in the current session/transaction
session.commit()

parent_id = Column(Integer, ForeignKey('parent.id'), unique=True)