python 缺少Greenlet:尚未调用greenlet_spawn

k7fdbhmy  于 2022-12-25  发布在  Python
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我尝试获取一对多关系中匹配的行数。当我尝试parent.children_count时,得到:
sqlalchemy.exc.MissingGreenlet:未调用greenlet_spawn;无法在此处调用await_only()。是否在意外位置尝试IO?(此错误的背景信息位于:https://sqlalche.me/e/14/xd2s
我添加了expire_on_commit=False,但仍然得到相同的错误。我该如何修复此错误?

import asyncio
from uuid import UUID, uuid4
from sqlmodel import SQLModel, Relationship, Field
from sqlalchemy.ext.asyncio import create_async_engine, AsyncSession

class Parent(SQLModel, table=True):
    id: UUID = Field(default_factory=uuid4, primary_key=True)
    children: list["Child"] = Relationship(back_populates="parent")
    @property
    def children_count(self):
        return len(self.children)

class Child(SQLModel, table=True):
    id: UUID = Field(default_factory=uuid4, primary_key=True)
    parent_id: UUID = Field(default=None, foreign_key=Parent.id)
    parent: "Parent" = Relationship(back_populates="children")

async def main():
    engine = create_async_engine("sqlite+aiosqlite://")
    async with engine.begin() as conn:
        await conn.run_sync(SQLModel.metadata.create_all)

    async with AsyncSession(engine) as session:
        parent = Parent()
        session.add(parent)
        await session.commit()
        await session.refresh(parent)
        print(parent.children_count)  # I expect 0 here, as of now this parent has no children

asyncio.run(main())
yizd12fk

yizd12fk1#

我认为这里的问题是默认情况下SQLAlchemy延迟加载关系,因此访问parent.children_count会隐式触发数据库查询,从而导致报告的错误。
解决此问题的一种方法是在关系定义中指定“惰性”以外的加载策略。使用SQLModel时,如下所示:

children: list['Child'] = Relationship(
    back_populates='parent', sa_relationship_kwargs={'lazy': 'selectin'}
)

这将导致SQLAlchemy发出额外的查询来获取关系,同时仍然处于“异步模式”。另一个选项是传递{'lazy': 'joined'},这将导致SQLAlchemy在单个JOIN查询中获取所有结果。
如果不需要配置关系,则可以发出指定以下选项的查询:

from sqlalchemy.orm import selectinload
from sqlmodel import select

    ...
    async with AsyncSession(engine) as session:
        parent = Parent()
        session.add(parent)
        await session.commit()
        result = await session.scalars(
            select(Parent).options(selectinload(Parent.children))
        )
        parent = result.first()
        print(
            parent.children_count
        )  # I need 0 here, as of now this parent has no children

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