Django REST Framework (DRF) 中“可调用默认函数”的正确使用

问题：

我怎样才能访问序列化器实例 or 任何相关论点 in the 默认可调用函数DRF 场？

设想：

我有一个序列化器配置如下，

def bar_value_callable_function(serializer_instance):
    if serializer_instance.context['request'].method == 'GET':
        return 'return value One'
    return 'return value Two'

类 FooSerializer(serializers.Serializer): 栏 = 序列化器.CharField(源='foo.bar'，默认=bar_value_callable_function)

Update
对于DRF>=3.12，使用这个默认类之王

class DefaultBarValue:
    requires_context = True

    def __call__(self, serializer_instance):
        if serializer_instance.context['request'].method == 'GET':
            return 'return value One'
        return 'return value Two'

pass a 类实例代替function to the default争论

# default callable class
class DefaultBarValue:
    """
    "Method `set_context` on defaults is deprecated and will
    no longer be called starting with 3.12. Instead set
    `requires_context = True` on the class, and accept the
    context as an additional argument.
    """
    requires_context = True  # for DRF>=3.12
    serializer_instance = None # not required for DRF>=3.12

    def set_context(self, serializer_instance): # not required for DRF>=3.12
        self.serializer_instance = serializer_instance

    def __call__(self, serializer_instance=None):
        if serializer_instance is None:  # will be None for older versions of DRF
            if self.serializer_instance.context['request'].method == 'GET':
                return 'return value One'
            return 'return value Two'
        else:  # for DRF>=3.12
            if serializer_instance.context['request'].method == 'GET':
                return 'return value One'
            return 'return value Two'

# serializer
class FooSerializer(serializers.Serializer):
    bar = serializers.CharField(source='foo.bar', default=DefaultBarValue())