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

在寻找解决方案时,我偶然发现了这个博客:https://blog.mosthege.net/2016/11/12/json-deserialization-of-nested-objects/

它使用与前面回答中相同的技术,但使用了装饰器。 我发现另一件有用的事情是,它在反序列化结束时返回一个类型化对象

class JsonConvert(object):
    class_mappings = {}

    @classmethod
    def class_mapper(cls, d):
        for keys, cls in clsself.mappings.items():
            if keys.issuperset(d.keys()):   # are all required arguments present?
                return cls(**d)
        else:
            # Raise exception instead of silently returning None
            raise ValueError('Unable to find a matching class for object: {!s}'.format(d))

    @classmethod
    def complex_handler(cls, Obj):
        if hasattr(Obj, '__dict__'):
            return Obj.__dict__
        else:
            raise TypeError('Object of type %s with value of %s is not JSON serializable' % (type(Obj), repr(Obj)))

    @classmethod
    def register(cls, claz):
        clsself.mappings[frozenset(tuple([attr for attr,val in cls().__dict__.items()]))] = cls
        return cls

    @classmethod
    def to_json(cls, obj):
        return json.dumps(obj.__dict__, default=cls.complex_handler, indent=4)

    @classmethod
    def from_json(cls, json_str):
        return json.loads(json_str, object_hook=cls.class_mapper)

用法:

@JsonConvert.register
class Employee(object):
    def __init__(self, Name:int=None, Age:int=None):
        self.Name = Name
        self.Age = Age
        return

@JsonConvert.register
class Company(object):
    def __init__(self, Name:str="", Employees:[Employee]=None):
        self.Name = Name
        self.Employees = [] if Employees is None else Employees
        return

company = Company("Contonso")
company.Employees.append(Employee("Werner", 38))
company.Employees.append(Employee("Mary"))

as_json = JsonConvert.to_json(company)
from_json = JsonConvert.from_json(as_json)
as_json_from_json = JsonConvert.to_json(from_json)

assert(as_json_from_json == as_json)

print(as_json_from_json)

其他回答

def load_model_from_dict(self, data: dict):
    for key, value in data.items():
        self.__dict__[key] = value
    return self

它帮助返回你自己的模型,从字典中不可预见的变量。

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。

因此,我正在寻找一种不需要大量自定义反序列化代码就能解组任意类型(想想数据类的字典,或者数据类数组的字典的字典)的方法。

这是我的方法:

import json
from dataclasses import dataclass, make_dataclass

from dataclasses_json import DataClassJsonMixin, dataclass_json


@dataclass_json
@dataclass
class Person:
    name: str


def unmarshal_json(data, t):
    Unmarhsal = make_dataclass('Unmarhsal', [('res', t)],
                               bases=(DataClassJsonMixin,))
    d = json.loads(data)
    out = Unmarhsal.from_dict({"res": d})
    return out.res


unmarshalled = unmarshal_json('{"1": {"name": "john"} }', dict[str, Person])
print(unmarshalled)

打印:{'1':Person(name='john')}

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。

使用json模块(Python 2.6新增)或几乎总是安装的simplejson模块。