我想指定一个模式字段,它接受一个或多个资源。但是,我似乎只能指定一种行为或另一种行为。
>>> class Resource(marshmallow.Schema):
... data = marshmallow.fields.Dict()
...
>>> class ContainerSchema(marshmallow.Schema):
... resource = marshmallow.fields.Nested(ResourceSchema, many=True)
...
>>> ContainerSchema().dump({'resource': [{'data': 'DATA'}]})
MarshalResult(data={'resource': [{'data': 'DATA'}]}, errors={})
在上面的示例中,必须定义一个列表。但是我不希望:
>>> ContainerSchema().dump({'resource': {'data': 'DATA'}})
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/lib64/python3.6/site-packages/marshmallow/schema.py", line 513, in dump
**kwargs
File "/lib64/python3.6/site-packages/marshmallow/marshalling.py", line 147, in serialize
index=(index if index_errors else None)
File "/lib64/python3.6/site-packages/marshmallow/marshalling.py", line 68, in call_and_store
value = getter_func(data)
File "/lib64/python3.6/site-packages/marshmallow/marshalling.py", line 141, in <lambda>
getter = lambda d: field_obj.serialize(attr_name, d, accessor=accessor)
File "/lib64/python3.6/site-packages/marshmallow/fields.py", line 252, in serialize
return self._serialize(value, attr, obj)
File "/lib64/python3.6/site-packages/marshmallow/fields.py", line 448, in _serialize
schema._update_fields(obj=nested_obj, many=self.many)
File "/lib64/python3.6/site-packages/marshmallow/schema.py", line 760, in _update_fields
ret = self.__filter_fields(field_names, obj, many=many)
File "/lib64/python3.6/site-packages/marshmallow/schema.py", line 810, in __filter_fields
obj_prototype = obj[0]
KeyError: 0
我可以有一个允许单个项目或多个项目的架构吗?