我有了自己的解决办法。使用此方法，将任何文档(字典、列表、ObjectId等)传递给序列化。

def getSerializable(doc):
    # check if it's a list
    if isinstance(doc, list):
        for i, val in enumerate(doc):
            doc[i] = getSerializable(doc[i])
        return doc

    # check if it's a dict
    if isinstance(doc, dict):
        for key in doc.keys():
            doc[key] = getSerializable(doc[key])
        return doc

    # Process ObjectId
    if isinstance(doc, ObjectId):
        doc = str(doc)
        return doc

    # Use any other custom serializting stuff here...

    # For the rest of stuff
    return doc

TLDR:复制-粘贴下面的选项1或选项2

真正的/完整的答案:让Pythons json模块与你的类一起工作

AKA，求解:json。dump ({"thing": YOUR_CLASS()})

解释:

Yes, a good reliable solution exists No, there is no python "official" solution By official solution, I mean there is no way (as of 2023) to add a method to your class (like toJSON in JavaScript) and/or no way to register your class with the built-in json module. When something like json.dumps([1,2, your_obj]) is executed, python doesn't check a lookup table or object method. I'm not sure why other answers don't explain this The closest official approach is probably andyhasit's answer which is to inherit from a dictionary. However, inheriting from a dictionary doesn't work very well for many custom classes like AdvancedDateTime, or pytorch tensors. The ideal workaround is this: Mutate json.dumps (affects everywhere, even pip modules that import json) Add def __json__(self) method to your class

选项1:让一个模块来做补丁

PIP安装json-fix (扩展+包装版FancyJohn的回答，谢谢@FancyJohn)

your_class_definition.py

import json_fix

class YOUR_CLASS:
    def __json__(self):
        # YOUR CUSTOM CODE HERE
        #    you probably just want to do:
        #        return self.__dict__
        return "a built-in object that is naturally json-able"

这是它。

使用示例:

from your_class_definition import YOUR_CLASS
import json

json.dumps([1,2, YOUR_CLASS()], indent=0)
# '[\n1,\n2,\n"a built-in object that is naturally json-able"\n]'

生成json。dump适用于Numpy数组，Pandas DataFrames和其他第三方对象，请参阅模块(只有大约2行代码，但需要解释)。

它是如何工作的?嗯…

选项2:补丁json。把你自己

注意:这种方法是简化的，它在已知的edgcase上失败(例如:如果你的自定义类继承了dict或其他内置类)，并且它错过了控制外部类的json行为(numpy数组，datetime, dataframes，张量等)。

some_file_thats_imported_before_your_class_definitions.py

# Step: 1
# create the patch
from json import JSONEncoder
def wrapped_default(self, obj):
    return getattr(obj.__class__, "__json__", wrapped_default.default)(obj)
wrapped_default.default = JSONEncoder().default
   
# apply the patch
JSONEncoder.original_default = JSONEncoder.default
JSONEncoder.default = wrapped_default

your_class_definition.py

# Step 2
class YOUR_CLASS:
    def __json__(self, **options):
        # YOUR CUSTOM CODE HERE
        #    you probably just want to do:
        #        return self.__dict__
        return "a built-in object that is natually json-able"

_

其他答案似乎都是“序列化自定义对象的最佳实践/方法”

在这里的文档中已经介绍过了(搜索“complex”可以找到编码复数的例子)

这个函数使用递归迭代遍历字典的每个部分，然后调用非内置类型类的repr()方法。

def sterilize(obj):
    object_type = type(obj)
    if isinstance(obj, dict):
        return {k: sterilize(v) for k, v in obj.items()}
    elif object_type in (list, tuple):
        return [sterilize(v) for v in obj]
    elif object_type in (str, int, bool, float):
        return obj
    else:
        return obj.__repr__()

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"
  }
}

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"
}

这是一个小库，它将一个对象及其所有子对象序列化为JSON，并将其解析回来:

https://github.com/tobiasholler/PyJSONSerialization/