我有一组表格如下:

workflows = Table('workflows', Base.metadata,
                  Column('id', Integer, primary_key=True),
                 )

actions = Table('actions', Base.metadata,
                Column('name', String, primary_key=True),
                Column('workflow_id', Integer, ForeignKey(workflows.c.id), primary_key=True),
               )

action_dependencies = Table('action_dependencies', Base.metadata,
                            Column('workflow_id', Integer, ForeignKey(workflows.c.id), primary_key=True),
                            Column('parent_action', String, ForeignKey(actions.c.name), primary_key=True),
                            Column('child_action', String, ForeignKey(actions.c.name), primary_key=True),
                           )

我的 ORM 类如下所示:

class Workflow(Base):
    __table__ = workflows

    actions = relationship("Action", order_by="Action.name", backref="workflow")


class Action(Base):
    __table__ = actions

    children = relationship("Action",
                            secondary=action_dependencies,
                            primaryjoin=actions.c.name == action_dependencies.c.parent_action,
                            secondaryjoin=actions.c.name == action_dependencies.c.child_action,
                            backref="parents"
                           )

因此在我的系统中，每个操作都由工作流 ID 及其名称的组合唯一标识。我希望每个 Action 都有 parents 和 children 属性来引用其父级和子级 Action 。每个 Action 可以有多个 parent 和 child 。

当我有一个函数时会出现问题，例如:

def set_parents(session, workflow_id, action_name, parents):
    action = session.query(db.Action).filter(db.Action.workflow_id == workflow.id).filter(db.Action.name == action_name).one()

    for parent_name in parents:
        parent = session.query(db.Action).filter(db.Action.workflow_id == workflow.id).filter(db.Action.name == parent_name).one()
        action.parents.append(parent)

    session.commit()

我收到如下错误:

IntegrityError: (IntegrityError) action_dependencies.workflow_id may not be NULL u'INSERT INTO action_dependencies (parent_action, child_action) VALUES (?, ?)' (u'directory_creator', u'packing')

如何获得正确设置 workflow_id 的关系？

最佳答案

请参阅下面的工作代码。关键点是我在评论中提到的那些:

• 适当的复合 ForeignKey
• 使用 FK 正确的关系配置

代码:

workflows = Table('workflows', Base.metadata,
                  Column('id', Integer, primary_key=True),
                 )

actions = Table('actions', Base.metadata,
                Column('workflow_id', Integer, ForeignKey(workflows.c.id), primary_key=True),
                Column('name', String, primary_key=True),
               )

action_dependencies = Table('action_dependencies', Base.metadata,
                            Column('workflow_id', Integer, ForeignKey(workflows.c.id), primary_key=True),
                            Column('parent_action', String, ForeignKey(actions.c.name), primary_key=True),
                            Column('child_action', String, ForeignKey(actions.c.name), primary_key=True),
                            ForeignKeyConstraint(['workflow_id', 'parent_action'], ['actions.workflow_id', 'actions.name']),
                            ForeignKeyConstraint(['workflow_id', 'child_action'], ['actions.workflow_id', 'actions.name']),
                           )
class Workflow(Base):
    __table__ = workflows
    actions = relationship("Action", order_by="Action.name", backref="workflow")

class Action(Base):
    __table__ = actions
    children = relationship("Action",
                            secondary=action_dependencies,
                            primaryjoin=and_(actions.c.name == action_dependencies.c.parent_action,
                                actions.c.workflow_id == action_dependencies.c.workflow_id),
                            secondaryjoin=and_(actions.c.name == action_dependencies.c.child_action,
                                actions.c.workflow_id == action_dependencies.c.workflow_id),
                            backref="parents"
                           )

# create db schema
Base.metadata.create_all(engine)

# create entities
w_1 = Workflow()
w_2 = Workflow()
a_11 = Action(name="ac-11", workflow=w_1)
a_12 = Action(name="ac-12", workflow=w_1)
a_21 = Action(name="ac-21", workflow=w_2)
a_22 = Action(name="ac-22", workflow=w_2)
session.add(w_1)
session.add(w_2)
a_22.parents.append(a_21)
session.commit()
session.expunge_all()
print '-'*80

# helper functions
def get_workflow(id):
    return session.query(Workflow).get(id)
def get_action(name):
    return session.query(Action).filter_by(name=name).one()

# test another OK
a_11 = get_action("ac-11")
a_12 = get_action("ac-12")
a_11.children.append(a_12)
session.commit()
session.expunge_all()
print '-'*80

# test KO (THIS SHOULD FAIL VIOLATING FK-constraint)
a_11 = get_action("ac-11")
a_22 = get_action("ac-22")
a_11.children.append(a_22)
session.commit()
session.expunge_all()
print '-'*80

关于python - SQLAlchemy:具有复合主键的关系表，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/10525797/