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

使用python 3.7,我发现下面的代码非常简单有效。在本例中,将JSON从文件加载到字典中:

class Characteristic:
    def __init__(self, characteristicName, characteristicUUID):
        self.characteristicName = characteristicName
        self.characteristicUUID = characteristicUUID


class Service:
    def __init__(self, serviceName, serviceUUID, characteristics):
        self.serviceName = serviceName
        self.serviceUUID = serviceUUID
        self.characteristics = characteristics

class Definitions:
    def __init__(self, services):
        self.services = []
        for service in services:
            self.services.append(Service(**service))


def main():
    parser = argparse.ArgumentParser(
        prog="BLEStructureGenerator",
        description="Taking in a JSON input file which lists all of the services, "
                    "characteristics and encoded properties. The encoding takes in "
                    "another optional template services and/or characteristics "
                    "file where the JSON file contents are applied to the templates.",
        epilog="Copyright Brown & Watson International"
    )

    parser.add_argument('definitionfile',
                        type=argparse.FileType('r', encoding='UTF-8'),
                        help="JSON file which contains the list of characteristics and "
                             "services in the required format")
    parser.add_argument('-s', '--services',
                        type=argparse.FileType('r', encoding='UTF-8'),
                        help="Services template file to be used for each service in the "
                             "JSON file list")
    parser.add_argument('-c', '--characteristics',
                        type=argparse.FileType('r', encoding='UTF-8'),
                        help="Characteristics template file to be used for each service in the "
                             "JSON file list")

    args = parser.parse_args()
    definition_dict = json.load(args.definitionfile)
    definitions = Definitions(**definition_dict)

其他回答

已经有多种可行的答案,但有一些由个人制作的小型库可以满足大多数用户的需求。

json2object就是一个例子。给定一个已定义的类,它将json数据反序列化到您的自定义模型,包括自定义属性和子对象。

它的使用非常简单。一个来自图书馆wiki的例子:

从json2object导入jsontoobject作为Jo 类学生: def __init__(自我): 自我。firstName =无 自我。lastName = None 自我。courses =[课程(")] 类课程: 定义__init__(self, name): Self.name = name 数据= " '{ “firstName”:“詹姆斯”, “姓”:“债券”, “课程”:[{ “名称”:“战斗”}, { “名称”:“射击”} ] } “‘ model = Student() Result = jo.deserialize(数据,模型) print (result.courses [0] . name)

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。

Dacite也可能是您的解决方案,它支持以下功能:

嵌套结构 (基本)类型检查 可选字段(即typing.Optional) 工会 向前引用 集合 自定义类型钩子

https://pypi.org/project/dacite/

from dataclasses import dataclass
from dacite import from_dict


@dataclass
class User:
    name: str
    age: int
    is_active: bool


data = {
    'name': 'John',
    'age': 30,
    'is_active': True,
}

user = from_dict(data_class=User, data=data)

assert user == User(name='John', age=30, is_active=True)

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')}

使用python 3.7,我发现下面的代码非常简单有效。在本例中,将JSON从文件加载到字典中:

class Characteristic:
    def __init__(self, characteristicName, characteristicUUID):
        self.characteristicName = characteristicName
        self.characteristicUUID = characteristicUUID


class Service:
    def __init__(self, serviceName, serviceUUID, characteristics):
        self.serviceName = serviceName
        self.serviceUUID = serviceUUID
        self.characteristics = characteristics

class Definitions:
    def __init__(self, services):
        self.services = []
        for service in services:
            self.services.append(Service(**service))


def main():
    parser = argparse.ArgumentParser(
        prog="BLEStructureGenerator",
        description="Taking in a JSON input file which lists all of the services, "
                    "characteristics and encoded properties. The encoding takes in "
                    "another optional template services and/or characteristics "
                    "file where the JSON file contents are applied to the templates.",
        epilog="Copyright Brown & Watson International"
    )

    parser.add_argument('definitionfile',
                        type=argparse.FileType('r', encoding='UTF-8'),
                        help="JSON file which contains the list of characteristics and "
                             "services in the required format")
    parser.add_argument('-s', '--services',
                        type=argparse.FileType('r', encoding='UTF-8'),
                        help="Services template file to be used for each service in the "
                             "JSON file list")
    parser.add_argument('-c', '--characteristics',
                        type=argparse.FileType('r', encoding='UTF-8'),
                        help="Characteristics template file to be used for each service in the "
                             "JSON file list")

    args = parser.parse_args()
    definition_dict = json.load(args.definitionfile)
    definitions = Definitions(**definition_dict)