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

修改@DS响应位,从一个文件加载:

def _json_object_hook(d): return namedtuple('X', d.keys())(*d.values())
def load_data(file_name):
  with open(file_name, 'r') as file_data:
    return file_data.read().replace('\n', '')
def json2obj(file_name): return json.loads(load_data(file_name), object_hook=_json_object_hook)

有一点:它不能加载前面有数字的项目。是这样的:

{
  "1_first_item": {
    "A": "1",
    "B": "2"
  }
}

因为“1_first_item”不是一个有效的python字段名。

其他回答

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

这里给出的答案没有返回正确的对象类型,因此我在下面创建了这些方法。如果你试图向给定JSON中不存在的类中添加更多字段,它们也会失败:

def dict_to_class(class_name: Any, dictionary: dict) -> Any:
    instance = class_name()
    for key in dictionary.keys():
        setattr(instance, key, dictionary[key])
    return instance


def json_to_class(class_name: Any, json_string: str) -> Any:
    dict_object = json.loads(json_string)
    return dict_to_class(class_name, dict_object)

如果你正在寻找将JSON或任何复杂字典的类型安全反序列化到python类中,我强烈推荐python 3.7+的pydantic。它不仅有一个简洁的API(不需要编写“helper”样板),可以与Python数据类集成,而且具有复杂和嵌套数据结构的静态和运行时类型验证。

使用示例:

from pydantic import BaseModel
from datetime import datetime

class Item(BaseModel):
    field1: str | int           # union
    field2: int | None = None   # optional
    field3: str = 'default'     # default values

class User(BaseModel):
    name: str | None = None
    username: str
    created: datetime           # default type converters
    items: list[Item] = []      # nested complex types

data = {
    'name': 'Jane Doe',
    'username': 'user1',
    'created': '2020-12-31T23:59:00+10:00',
    'items': [
        {'field1': 1, 'field2': 2},
        {'field1': 'b'},
        {'field1': 'c', 'field3': 'override'}
    ]
}

user: User = User(**data)

要了解更多细节和特性,请查看文档中的pydantic的rational部分。

这里有一个快速而肮脏的json pickle替代方案

import json

class User:
    def __init__(self, name, username):
        self.name = name
        self.username = username

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

    @classmethod
    def from_json(cls, json_str):
        json_dict = json.loads(json_str)
        return cls(**json_dict)

# example usage
User("tbrown", "Tom Brown").to_json()
User.from_json(User("tbrown", "Tom Brown").to_json()).to_json()

你可以试试这个:

class User(object):
    def __init__(self, name, username):
        self.name = name
        self.username = username

import json
j = json.loads(your_json)
u = User(**j)

只需创建一个新对象,并将参数作为映射传递。


你也可以有一个带有对象的JSON:

import json
class Address(object):
    def __init__(self, street, number):
        self.street = street
        self.number = number

    def __str__(self):
        return "{0} {1}".format(self.street, self.number)

class User(object):
    def __init__(self, name, address):
        self.name = name
        self.address = Address(**address)

    def __str__(self):
        return "{0} ,{1}".format(self.name, self.address)

if __name__ == '__main__':
    js = '''{"name":"Cristian", "address":{"street":"Sesame","number":122}}'''
    j = json.loads(js)
    print(j)
    u = User(**j)
    print(u)