我想将JSON数据转换为Python对象。

我从Facebook API收到JSON数据对象,我想将其存储在数据库中。

我的当前视图在Django (Python)(请求。POST包含JSON):

response = request.POST
user = FbApiUser(user_id = response['id'])
user.name = response['name']
user.username = response['username']
user.save()

这很好,但是如何处理复杂的JSON数据对象呢? 如果我能以某种方式将这个JSON对象转换为易于使用的Python对象,是不是会更好?


当前回答

如果你使用的是Python 3.5+,你可以使用json来序列化和反序列化到普通的旧Python对象:

import jsons

response = request.POST

# You'll need your class attributes to match your dict keys, so in your case do:
response['id'] = response.pop('user_id')

# Then you can load that dict into your class:
user = jsons.load(response, FbApiUser)

user.save()

你也可以让FbApiUser从jsons继承。JsonSerializable更优雅:

user = FbApiUser.from_json(response)

如果你的类由Python默认类型组成,比如字符串、整数、列表、日期时间等,这些例子就可以工作。不过,jsons lib需要自定义类型的类型提示。

其他回答

JSON到python对象

下面的代码递归地使用对象键创建动态属性。

JSON对象- fb_data.json:

{
    "name": "John Smith",
    "hometown": {
        "name": "New York",
        "id": 123
    },
    "list": [
        "a",
        "b",
        "c",
        1,
        {
            "key": 1
        }
    ],
    "object": {
        "key": {
            "key": 1
        }
    }
}

在转换中我们有三种情况:

列表 Dicts(新对象) Bool, int, float和STR

import json


class AppConfiguration(object):
    def __init__(self, data=None):
        if data is None:
            with open("fb_data.json") as fh:
                data = json.loads(fh.read())
        else:
            data = dict(data)

        for key, val in data.items():
            setattr(self, key, self.compute_attr_value(val))

    def compute_attr_value(self, value):
        if isinstance(value, list):
            return [self.compute_attr_value(x) for x in value]
        elif isinstance(value, dict):
            return AppConfiguration(value)
        else:
            return value


if __name__ == "__main__":
    instance = AppConfiguration()

    print(instance.name)
    print(instance.hometown.name)
    print(instance.hometown.id)
    print(instance.list[4].key)
    print(instance.object.key.key)

键值对是属性-对象。

输出:

John Smith
New York
123
1
1

将JSON作为代码粘贴

支持TypeScript、Python、Go、Ruby、c#、Java、Swift、Rust、Kotlin、c++、Flow、Objective-C、JavaScript、Elm、JSON Schema。

从JSON、JSON Schema和TypeScript中交互式地生成类型和(反)序列化代码 将JSON/JSON Schema/TypeScript作为代码粘贴

quicktype从示例JSON数据中推断类型,然后输出强类型模型和序列化器,以便用所需的编程语言处理这些数据。

输出:

# Generated by https://quicktype.io
#
# To change quicktype's target language, run command:
#
#   "Set quicktype target language"

from typing import List, Union


class Hometown:
    name: str
    id: int

    def __init__(self, name: str, id: int) -> None:
        self.name = name
        self.id = id


class Key:
    key: int

    def __init__(self, key: int) -> None:
        self.key = key


class Object:
    key: Key

    def __init__(self, key: Key) -> None:
        self.key = key


class FbData:
    name: str
    hometown: Hometown
    list: List[Union[Key, int, str]]
    object: Object

    def __init__(self, name: str, hometown: Hometown, list: List[Union[Key, int, str]], object: Object) -> None:
        self.name = name
        self.hometown = hometown
        self.list = list
        self.object = object

这个扩展可以在Visual Studio代码市场中免费获得。

如果你正在使用python 3.6+,你可以使用棉花糖-数据类。与上面列出的所有解决方案相反,它既简单,又类型安全:

from marshmallow_dataclass import dataclass

@dataclass
class User:
    name: str

user = User.Schema().load({"name": "Ramirez"})

我认为最简单的解决方法是

import orjson  # faster then json =)
from typing import NamedTuple

_j = '{"name":"Иван","age":37,"mother":{"name":"Ольга","age":58},"children":["Маша","Игорь","Таня"],"married": true,' \
     '"dog":null} '


class PersonNameAge(NamedTuple):
    name: str
    age: int


class UserInfo(NamedTuple):
    name: str
    age: int
    mother: PersonNameAge
    children: list
    married: bool
    dog: str


j = orjson.loads(_j)
u = UserInfo(**j)

print(u.name, u.age, u.mother, u.children, u.married, u.dog)

>>> Ivan 37 {'name': 'Olga', 'age': 58} ['Mary', 'Igor', 'Jane'] True None

既然没有人给出了和我一样的答案,我就把它贴在这里。

这是一个健壮的类,可以轻松地在JSON str和dict之间来回转换,我已经从我的答案复制到另一个问题:

import json

class PyJSON(object):
    def __init__(self, d):
        if type(d) is str:
            d = json.loads(d)

        self.from_dict(d)

    def from_dict(self, d):
        self.__dict__ = {}
        for key, value in d.items():
            if type(value) is dict:
                value = PyJSON(value)
            self.__dict__[key] = value

    def to_dict(self):
        d = {}
        for key, value in self.__dict__.items():
            if type(value) is PyJSON:
                value = value.to_dict()
            d[key] = value
        return d

    def __repr__(self):
        return str(self.to_dict())

    def __setitem__(self, key, value):
        self.__dict__[key] = value

    def __getitem__(self, key):
        return self.__dict__[key]

json_str = """... JSON string ..."""

py_json = PyJSON(json_str)

你可以使用

x = Map(json.loads(response))
x.__class__ = MyClass

在哪里

class Map(dict):
    def __init__(self, *args, **kwargs):
        super(Map, self).__init__(*args, **kwargs)
        for arg in args:
            if isinstance(arg, dict):
                for k, v in arg.iteritems():
                    self[k] = v
                    if isinstance(v, dict):
                        self[k] = Map(v)

        if kwargs:
            # for python 3 use kwargs.items()
            for k, v in kwargs.iteritems():
                self[k] = v
                if isinstance(v, dict):
                    self[k] = Map(v)

    def __getattr__(self, attr):
        return self.get(attr)

    def __setattr__(self, key, value):
        self.__setitem__(key, value)

    def __setitem__(self, key, value):
        super(Map, self).__setitem__(key, value)
        self.__dict__.update({key: value})

    def __delattr__(self, item):
        self.__delitem__(item)

    def __delitem__(self, key):
        super(Map, self).__delitem__(key)
        del self.__dict__[key]

对于通用的、经得起未来考验的解决方案。