python - 如何序列化 umongo 文档

Question

假设我有以下示例：

from datetime import datetime
import dill
from pymongo import MongoClient
from umongo import Instance, Document, fields, validate

db = MongoClient().test
instance = Instance(db)

@instance.register
class User(Document):
    email = fields.EmailField(required=True, unique=True)
    birthday = fields.DateTimeField(validate=validate.Range(min=datetime(1900, 1, 1)))
    friends = fields.ListField(fields.ReferenceField("User"))

    class Meta:
        collection = db.user

User.ensure_indexes()

goku = User(email='[e-mail]', birthday=datetime(1984, 11, 20))
goku.commit()

found_goku = User.find_one({"email": '[e-mail]'})
with open("dumped.dil", "wb") as out_file:
    dill.dump(found_goku, out_file)

这会导致以下错误dill.dump：

_pickle.PicklingError: Can't pickle <Implementation class 'tests.test_serialize_umongo.User'>: it's not found as tests.test_serialize_umongo.User

我知道我不能简单地腌制用户，因为我不能腌制数据库实例。我想使用 umongodump和覆盖将用户腌制到字典__getstate__，但我不知道如何解决这个it's not found as错误。

标签： pythonserializationpickleumongo

Answer 1

如果我理解正确，我相信您正在寻找的是以下内容。本质上，我们使用文档自己的dump()方法将序列化的文档作为要腌制的用户传递。这里唯一的区别是最后一行dill.dump(found_goku.dump(), out_file)：：

from datetime import datetime
import dill
from pymongo import MongoClient
from umongo import Instance, Document, fields, validate

db = MongoClient().test
instance = Instance(db)

@instance.register
class User(Document):
    email = fields.EmailField(required=True, unique=True)
    birthday = fields.DateTimeField(validate=validate.Range(min=datetime(1900, 1, 1)))
    friends = fields.ListField(fields.ReferenceField("User"))

    class Meta:
        collection = db.user

User.ensure_indexes()

goku = User(email='[e-mail]', birthday=datetime(1984, 11, 20))
goku.commit()

found_goku = User.find_one({"email": '[e-mail]'})
with open("dumped.dil", "wb") as out_file:
    dill.dump(found_goku.dump(), out_file)