一个非常简单的一行程序解决方案

import json

json.dumps(your_object, default=lambda __o: __o.__dict__)

结束!

下面是一个测试。

import json
from dataclasses import dataclass


@dataclass
class Company:
    id: int
    name: str

@dataclass
class User:
    id: int
    name: str
    email: str
    company: Company


company = Company(id=1, name="Example Ltd")
user = User(id=1, name="John Doe", email="john@doe.net", company=company)


json.dumps(user, default=lambda __o: __o.__dict__)

输出:

{
  "id": 1, 
  "name": "John Doe", 
  "email": "john@doe.net", 
  "company": {
    "id": 1, 
    "name": "Example Ltd"
  }
}

对于更复杂的类，您可以考虑使用jsonpickle工具:

jsonpickle is a Python library for serialization and deserialization of complex Python objects to and from JSON. The standard Python libraries for encoding Python into JSON, such as the stdlib’s json, simplejson, and demjson, can only handle Python primitives that have a direct JSON equivalent (e.g. dicts, lists, strings, ints, etc.). jsonpickle builds on top of these libraries and allows more complex data structures to be serialized to JSON. jsonpickle is highly configurable and extendable–allowing the user to choose the JSON backend and add additional backends.

(链接到PyPi上的jsonpickle)

这对我来说很有效:

class JsonSerializable(object):

    def serialize(self):
        return json.dumps(self.__dict__)

    def __repr__(self):
        return self.serialize()

    @staticmethod
    def dumper(obj):
        if "serialize" in dir(obj):
            return obj.serialize()

        return obj.__dict__

然后

class FileItem(JsonSerializable):
    ...

and

log.debug(json.dumps(<my object>, default=JsonSerializable.dumper, indent=2))

下面是一个简单功能的简单解决方案:

.toJSON()方法

实现一个序列化器方法，而不是一个JSON可序列化类:

import json

class Object:
    def toJSON(self):
        return json.dumps(self, default=lambda o: o.__dict__, 
            sort_keys=True, indent=4)

所以你只需调用它来序列化:

me = Object()
me.name = "Onur"
me.age = 35
me.dog = Object()
me.dog.name = "Apollo"

print(me.toJSON())

将输出:

{
    "age": 35,
    "dog": {
        "name": "Apollo"
    },
    "name": "Onur"
}

如果你正在使用Python3.5+，你可以使用jsons。(PyPi: https://pypi.org/project/jsons/)它将把你的对象(及其所有属性递归地)转换为字典。

import jsons

a_dict = jsons.dump(your_object)

或者如果你想要一个字符串:

a_str = jsons.dumps(your_object)

或者你的类实现了jsons。JsonSerializable:

a_dict = your_object.json

如果你能够安装一个软件包，我建议你试试dill，它在我的项目中工作得很好。这个包的一个优点是它具有与pickle相同的接口，因此如果您已经在项目中使用了pickle，则可以简单地替换为dill并查看脚本是否运行，而无需更改任何代码。所以这是一个非常便宜的解决方案!

(完全反披露:我与莳萝项目没有任何关联，也从未参与过。)

安装包:

pip install dill

然后编辑你的代码导入莳萝而不是pickle:

# import pickle
import dill as pickle

运行脚本，看看它是否有效。(如果是的话，你可能想要清理你的代码，这样你就不再隐藏pickle模块的名字了!)

关于dill可以和不能序列化的数据类型的一些细节，来自项目页面:

dill can pickle the following standard types: none, type, bool, int, long, float, complex, str, unicode, tuple, list, dict, file, buffer, builtin, both old and new style classes, instances of old and new style classes, set, frozenset, array, functions, exceptions dill can also pickle more ‘exotic’ standard types: functions with yields, nested functions, lambdas, cell, method, unboundmethod, module, code, methodwrapper, dictproxy, methoddescriptor, getsetdescriptor, memberdescriptor, wrapperdescriptor, xrange, slice, notimplemented, ellipsis, quit dill cannot yet pickle these standard types: frame, generator, traceback