python - SQLALchemy session.merge 不起作用

Question

AFAIK 合并执行并 insert or update所以我正在做的事情非常简单。集线器和标签之间存在 1 到 N 的关系。因此，当我尝试在标签和集线器上进行合并时，集线器运行良好并从数据库加载现有集线器并进行更新，但在 db.session.merge(nuevo_tag) 时失败执行时会抛出异常，因为即使标签先前存在，也会在幕后尝试进行插入。我做错了什么？

nuevo_hub = Hub(guid_hub,name,location,comments,id_project,creado_en,actualizado_en)
merged_hub = db.session.merge(nuevo_hub)
#db.session.commit() # If I use this line tags perform an insert.
nuevo_tag = Tag(guid_tag,project,merged_hub,TYPE_HUB,creado_en,actualizado_en)
merged_tag = db.session.merge(nuevo_tag)
db.session.commit()

如果我删除 db.session.commit()然后显示其他错误:

sqlalchemy.orm.exc.FlushError: New instance with identity key (, (b'\x11\x0e\x84\x00\xe2\x9b\x11\xd4\xa7\x16DfUD\x00\r',)) conflicts with persistent instance

class Item(db.Model):
    __tablename__ = "items"
    # id_item     = db.Column(db.Integer, autoincrement=True, primary_key=True)
    guid_item   = db.Column(db.BINARY(16), primary_key=True)
    id_project  = db.Column(db.Integer,db.ForeignKey("projects.id_project"))
    type        = db.Column(db.Integer)
    name        = db.Column(db.String(50), nullable=False, index= True)
    created_at  = db.Column(db.DateTime)
    updated_at  = db.Column(db.DateTime)
    __mapper_args__ = {
        'polymorphic_identity': 'items',
        'polymorphic_on':type,
        'with_polymorphic':'*'
    }
    __table_args__ = (
                        db.UniqueConstraint('name', 'id_project', name='_unique_name_project'),
                     )

    def __init__(self,creado_en=None):
        self.created_at = creado_en
        self.updated_at = creado_en


class Hub(Item):
    __tablename__ = "hubs"
    __mapper_args__ = {
        'polymorphic_identity': TYPE_HUB,
        'with_polymorphic':'*'
    }
    guid_hub = db.Column(db.BINARY(16), db.ForeignKey(Item.guid_item), primary_key=True)
    location = db.Column(db.String(50))
    comments = db.Column(db.String(128))

    def __init__(self, guid_hub=None, nombre=None, location=None,comments=None, id_project=None, creado_en=None, actualizado_en=None):
        self.type = TYPE_HUB
        self.guid_item = guid_hub
        self.guid_hub = guid_hub
        self.name = nombre
        self.id_project = id_project
        self.location = location
        self.comments = comments
        self.created_at = creado_en
        self.updated_at = actualizado_en


class Tag(db.Model):
    __tablename__ = "tags"
    guid_tag      = db.Column(db.BINARY(16), primary_key=True)
    id_project  = db.Column(db.Integer,db.ForeignKey("projects.id_project"))
    guid_item     = db.Column(db.BINARY(16),db.ForeignKey("items.guid_item"))
    project     = db.relationship(Proyecto, backref=db.backref('list_tags', lazy='dynamic'))
    item        = db.relationship(Item, backref=db.backref('list_tags', lazy='joined'))
    type        = db.Column(db.Integer) #(0,hub);(1,cable);(2,pipe);(3,electrical_pipes)
    created_at  = db.Column(db.DateTime)
    updated_at  = db.Column(db.DateTime)

    def __init__(self,guid_tag,project,item,type,created_at,updated_at):
        # self.guid_item = guid_tag
        self.guid_tag = guid_tag
        self.project = project
        self.item = item
        self.type = type
        self.created_at = created_at
        self.updated_at = updated_at

Answer 1

我在 docs 中找到了答案。考虑 Hub 是 Item 的子类。我必须将延迟引用从连接更改为动态。

class Tag(db.Model):
    __tablename__ = "tags"
    guid_tag      = db.Column(db.BINARY(16), primary_key=True)
    id_project  = db.Column(db.Integer,db.ForeignKey("projects.id_project"))
    guid_item     = db.Column(db.BINARY(16),db.ForeignKey("items.guid_item"))
    project     = db.relationship(Proyecto, backref=db.backref('list_tags', lazy='dynamic'))
    item        = db.relationship(Item, backref=db.backref('list_tags', lazy='joined'))
    type        = db.Column(db.Integer) #(0,hub);(1,cable);(2,pipe);(3,electrical_pipes)
    created_at  = db.Column(db.DateTime)
    updated_at  = db.Column(db.DateTime)

换句话说，我必须避免在同一个 session 中拥有持久对象的两个副本。