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?
你可以使用
x = Map(json.loads(response))
x.__class__ = MyClass
在哪里
class Map(dict):
def __init__(self, *args, **kwargs):
super(Map, self).__init__(*args, **kwargs)
for arg in args:
if isinstance(arg, dict):
for k, v in arg.iteritems():
self[k] = v
if isinstance(v, dict):
self[k] = Map(v)
if kwargs:
# for python 3 use kwargs.items()
for k, v in kwargs.iteritems():
self[k] = v
if isinstance(v, dict):
self[k] = Map(v)
def __getattr__(self, attr):
return self.get(attr)
def __setattr__(self, key, value):
self.__setitem__(key, value)
def __setitem__(self, key, value):
super(Map, self).__setitem__(key, value)
self.__dict__.update({key: value})
def __delattr__(self, item):
self.__delitem__(item)
def __delitem__(self, key):
super(Map, self).__delitem__(key)
del self.__dict__[key]
对于通用的、经得起未来考验的解决方案。
你可以试试这个:
class User(object):
def __init__(self, name, username):
self.name = name
self.username = username
import json
j = json.loads(your_json)
u = User(**j)
只需创建一个新对象,并将参数作为映射传递。
你也可以有一个带有对象的JSON:
import json
class Address(object):
def __init__(self, street, number):
self.street = street
self.number = number
def __str__(self):
return "{0} {1}".format(self.street, self.number)
class User(object):
def __init__(self, name, address):
self.name = name
self.address = Address(**address)
def __str__(self):
return "{0} ,{1}".format(self.name, self.address)
if __name__ == '__main__':
js = '''{"name":"Cristian", "address":{"street":"Sesame","number":122}}'''
j = json.loads(js)
print(j)
u = User(**j)
print(u)
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)
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)
这是我的办法。
特性
支持类型提示
如果缺少键则引发错误。
跳过数据中的额外值
import typing
class User:
name: str
age: int
def __init__(self, data: dict):
for k, _ in typing.get_type_hints(self).items():
setattr(self, k, data[k])
data = {
"name": "Susan",
"age": 18
}
user = User(data)
print(user.name, user.age)
# Output: Susan 18