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
数据类向导是一种现代的选项，可以类似地为您工作。它支持自动键大小写转换，如camelCase或TitleCase，这两者在API响应中都很常见。
当将实例转储到dict/JSON时，默认的键转换是camelCase，但这可以很容易地使用主数据类上提供的Meta配置来覆盖。
https://pypi.org/project/dataclass-wizard/
from dataclasses import dataclass
from dataclass_wizard import fromdict, asdict
@dataclass
class User:
name: str
age: int
is_active: bool
data = {
'name': 'John',
'age': 30,
'isActive': True,
}
user = fromdict(User, data)
assert user == User(name='John', age=30, is_active=True)
json_dict = asdict(user)
assert json_dict == {'name': 'John', 'age': 30, 'isActive': True}
设置元配置的例子，当序列化为dict/JSON时，将字段转换为lisp-case:
DumpMeta(key_transform='LISP').bind_to(User)
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
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)
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)
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。
使用json模块(Python 2.6新增)或几乎总是安装的simplejson模块。
