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?
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')}
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
更新
在Python3中,你可以使用SimpleNamespace和object_hook在一行中完成:
import json
from types import SimpleNamespace
data = '{"name": "John Smith", "hometown": {"name": "New York", "id": 123}}'
# Parse JSON into an object with attributes corresponding to dict keys.
x = json.loads(data, object_hook=lambda d: SimpleNamespace(**d))
print(x.name, x.hometown.name, x.hometown.id)
旧答案(Python2)
在Python2中,你可以使用namedtuple和object_hook在一行中完成(但对于嵌套对象非常慢):
import json
from collections import namedtuple
data = '{"name": "John Smith", "hometown": {"name": "New York", "id": 123}}'
# Parse JSON into an object with attributes corresponding to dict keys.
x = json.loads(data, object_hook=lambda d: namedtuple('X', d.keys())(*d.values()))
print x.name, x.hometown.name, x.hometown.id
或者,为了便于重用:
def _json_object_hook(d): return namedtuple('X', d.keys())(*d.values())
def json2obj(data): return json.loads(data, object_hook=_json_object_hook)
x = json2obj(data)
如果希望它处理不是很好的属性名称的键,请检查namedtuple的rename参数。
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