从Python Marshmallow中的兄弟字段访问反序列化的数据?

时间:2015-12-23 09:51:59

标签: python json serialization marshmallow

我有以下Python类和Marshmallow架构定义:

from marshmallow import Schema, fields, post_load

class Author:
    def __init__(self, id, name):
        self.id = id
        self.name = name


class Book:
    def __init__(self, id, author, name):
        self.id = id
        assert isinstance(author, Author)
        self.author = author
        self.name = name


class Library:
    def __init__(self, authors, books):
        self.authors = authors
        self.books = books


class AuthorSchema(Schema):
    id = fields.Int()
    name = fields.Str()

    @post_load
    def make_obj(self, data):
        return Author(**data)


class BookSchema(Schema):
    id = fields.Int()
    author_id = fields.Method('get_id', 'get_author', attribute="author")
    name = fields.Str()

    @post_load
    def make_obj(self, data):
        return Book(**data)

    def get_id(self, obj):
        return obj.author.id

    def get_author(self, value):
        return [a for a in authors if a.id == value][0]


class LibrarySchema(Schema):
    authors = fields.List(fields.Nested(AuthorSchema()))
    books = fields.List(fields.Nested(BookSchema()))

    @post_load
    def make_obj(self, data):
        return Library(**data)

    # preserve ordering of fields
    class Meta:
        ordered = True


def test_author_referencing():
    author1 = Author(1, "Astrid Lindgren")
    author2 = Author(2, "Tove Jansson")

    book1 = Book(11, author1, "The Brothers Lionheart")
    book2 = Book(12, author2, "Comet in Moominland")

    library = Library(authors=[author1, author2], books=[book1, book2])

    schema = LibrarySchema(strict=True)

    library_dict = schema.dump(library).data
    library2 = schema.load(library_dict).data

正如希望的样本所示,我希望有一个数据模型,其中书籍对象包含对作者的引用(而不仅仅是作者ID),但是会被序列化为作者ID。

显然,序列化不是问题,但在反序列化时, 我需要访问作者列表中已经反序列化的内容。我不太清楚如何用Marshmallow做到这一点。它甚至可能吗?当然,在这种情况下,我可以使用数字作者ID实例化Book个对象,并在@post_load中执行LibrarySchema操作,以使用引用替换作者ID,但这样做会让人感到肮脏和笨拙我。请帮忙。 : - )

1 个答案:

答案 0 :(得分:0)

在post_load中链接数据正是你应该如何做到的。