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?
改进lovasoa非常好的答案。
如果你正在使用python 3.6+,你可以使用:
PIP安装棉花糖-enum和
PIP安装棉花糖数据类
它简单且类型安全。
你可以在string-json中转换你的类,反之亦然:
从对象到字符串Json:
from marshmallow_dataclass import dataclass
user = User("Danilo","50","RedBull",15,OrderStatus.CREATED)
user_json = User.Schema().dumps(user)
user_json_str = user_json.data
从String Json到Object:
json_str = '{"name":"Danilo", "orderId":"50", "productName":"RedBull", "quantity":15, "status":"Created"}'
user, err = User.Schema().loads(json_str)
print(user,flush=True)
类定义:
class OrderStatus(Enum):
CREATED = 'Created'
PENDING = 'Pending'
CONFIRMED = 'Confirmed'
FAILED = 'Failed'
@dataclass
class User:
def __init__(self, name, orderId, productName, quantity, status):
self.name = name
self.orderId = orderId
self.productName = productName
self.quantity = quantity
self.status = status
name: str
orderId: str
productName: str
quantity: int
status: OrderStatus
查看JSON模块文档中的专门化JSON对象解码一节。您可以使用它将JSON对象解码为特定的Python类型。
这里有一个例子:
class User(object):
def __init__(self, name, username):
self.name = name
self.username = username
import json
def object_decoder(obj):
if '__type__' in obj and obj['__type__'] == 'User':
return User(obj['name'], obj['username'])
return obj
json.loads('{"__type__": "User", "name": "John Smith", "username": "jsmith"}',
object_hook=object_decoder)
print type(User) # -> <type 'type'>
更新
如果你想通过json模块访问字典中的数据,可以这样做:
user = json.loads('{"__type__": "User", "name": "John Smith", "username": "jsmith"}')
print user['name']
print user['username']
就像一本普通的字典。
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')}
这不是代码高尔夫,但这里是我使用类型的最短技巧。SimpleNamespace作为JSON对象的容器。
与namedtuple解决方案相比,它是:
可能更快/更小,因为它没有为每个对象创建一个类
更短的
没有重命名选项,对于不是有效标识符的键可能有相同的限制(在幕后使用setattr)
例子:
from __future__ import print_function
import json
try:
from types import SimpleNamespace as Namespace
except ImportError:
# Python 2.x fallback
from argparse import Namespace
data = '{"name": "John Smith", "hometown": {"name": "New York", "id": 123}}'
x = json.loads(data, object_hook=lambda d: Namespace(**d))
print (x.name, x.hometown.name, x.hometown.id)
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)