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我想序列化一个包含嵌套列表的 python 列表。下面的代码构造了要从 gnome 密钥环序列化的对象，但jsonpickle编码器不会序列化子列表。有了unpicklable=True，我简单地得到：

[{"py/object": "__main__.Collection", "label": ""}, {"py/object": "__main__.Collection", "label": "Login"}]

我已经尝试过设置/不设置max_depth并尝试了很多深度数字，但无论如何，腌制器只会腌制顶级物品。

如何让它序列化整个对象结构？

#! /usr/bin/env python

import secretstorage
import jsonpickle

class Secret(object):
    label = ""
    username = ""
    password = ""

    def __init__(self, secret):
        self.label = secret.get_label()
        self.password = '%s' % secret.get_secret()
        attributes = secret.get_attributes()
        if attributes and 'username_value' in attributes:
            self.username = '%s' % attributes['username_value']

class Collection(object):
    label = ""
    secrets = []

    def __init__(self, collection):
        self.label = collection.get_label()
        for secret in collection.get_all_items():
            self.secrets.append(Secret(secret))


def keyring_to_json():
    collections = []
    bus = secretstorage.dbus_init()
    for collection in secretstorage.get_all_collections(bus):
        collections.append(Collection(collection))

    pickle = jsonpickle.encode(collections, unpicklable=False);
    print(pickle)


if __name__ == '__main__':
    keyring_to_json()
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1 回答 1

2

我遇到了同样的问题，并且能够通过在 init 中移动数组的声明来解决它：

class Collection(object):
    label = ""
    # secrets = [] (move this into __init__)

   def __init__(self, collection):
        self.secrets = []
        self.label = collection.get_label()
        for secret in collection.get_all_items():
            self.secrets.append(Secret(secret))
于 2016-03-06T19:22:59.707 回答