如何在不带"_"的情况下将具有属性的对象转换为JSON在Python 3中? [英] How to convert Object with Properties to JSON without "_" in Python 3?

问题描述

我想将Python对象转换为JSON格式.

I would like to convert an Python object into JSON-format.

User 类的私有属性是使用属性定义的.我在此处找到方法 to_Json()

The private attributes of the class User are defined using properties. The method to_Json() I have found here

class User:
  def __init__(self):
      self._name = None
      self._gender = None

  @property    
  def name(self):
    return self._name     

  @name.setter    
  def name(self, name):
      self._name = name

  @property    
  def gender(self):
      return self._gender     

  @gender.setter    
  def gender(self, gender):
    self._gender = gender

  def to_Json(self):
      return json.dumps(self, default=lambda o: o.__dict__, allow_nan=False, sort_keys=False, indent=4)

使用此类和方法的输出为:

The output using this class and method is:

{
  "_name": "Peter",
  "_age": 26
}

摆脱JSON格式下划线的最简单方法是什么? (我希望使用名称"而不是"_name".)由于我遇到错误(最大递归深度)，因此无法删除类中的下划线.我认为重命名属性的方法可以解决此问题，但这是最好的解决方案吗?

What is the easiest way to get rid of the underscores in the JSON-format? (I want "name" instead of "_name") Removing the underscore in the class is not an option since I get an error (max. recursion depth). I think renaming the methods of the attributes would solve this problem, but is this the best solution here?

重命名json.dumbs之前的所有键(请参见此处)是不切实际的方法，因为我的课比上面的例子还要复杂.

Renaming all keys before the json.dumbs (see here) is not a practical approach because I my class is more complex than the above example.

那么，将Python对象尽快转换为JSON格式的最佳实践是什么?

So, what is the best practice to convert a Python object into JSON-format as fast as possible?

推荐答案

如果您发布的示例代码反映了您的真实代码，则根本没有任何理由使用该属性.您可以这样做:

If the example code you posted mirrors your real code, there really isn't any reason for the properties at all. You could just do:

class User(object):
    def __init__(self):
        self.name = None
        self.age = None

因为您并没有真正在底线和属性后面向用户隐藏任何内容.

since you're not really hiding anything from the user behind the underscores and properties anyway.

如果您要做需要进行转换，我想在自定义编码器中进行:

If you do need to do the transformation, I like to do it in a custom encoder:

class MyEncoder(json.JSONEncoder):
    def default(self, o):
        return {k.lstrip('_'): v for k, v in vars(o).items()}

json_encoded_user = json.dumps(some_user, cls=MyEncoder)

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