Question

我有数据我正试图进入数据库，它包含一个我想用来为另一个字段生成数据的字段。我完全混淆了如何让我的序列化程序忽略但仍然使用obj.owner_name来获取owner_id用于PUT。

如果有人能指出我做错了什么就会很棒。

<root>
  <list-item>
    <varmst_id>1867</varmst_id>
    <varmst_type>3</varmst_type>
    <varmst_name>var_date                      </varmst_name>
    <varmst_value>20140911</varmst_value>
    <varmst_desc/>
    <varmst_public>Y</varmst_public>
    <owner_name>Operations</owner_name>
  </list-item>
</root>

这是我的序列化程序 -

class VariablePUTSerializer(serializers.ModelSerializer):
    owner_id = serializers.SerializerMethodField('get_owner_id')

    class Meta:
        model = Varmst
        resource_name = 'varmst'
        fields = ('varmst_id', 'varmst_type', 'varmst_name', 'varmst_value', 'varmst_desc',
                'varmst_public', 'owner_id')
        exclude_fields = ('owner_name',)

    def transform_varmst_id(self, obj, value):
        maxid = Varmst.objects.latest('varmst_id').varmst_id
        if Varmst.objects.filter(varmst_name=obj.varmst_name).exists():
            obj.varmst_id = Varmst.objects.filter(varmst_name=obj.varmst_name).values_list('varmst_id')[0]
            return obj.varmst_id
        else:
            obj.varmst_id = maxid + 1
            return obj.varmst_id

    def get_owner_id(self, obj):
        obj.owner_id = Owner.objects.filter(owner_name=obj.owner_name).values_list('owner_id')[0]
        return obj.owner_id

当我尝试使用上面的数据进行PUT时，这是我的追溯 -

Traceback:
File "D:\Python27\lib\site-packages\django\core\handlers\base.py" in get_response
  112.                     response = wrapped_callback(request, *callback_args, **callback_kwargs)
File "D:\Python27\lib\site-packages\django\views\generic\base.py" in view
  69.             return self.dispatch(request, *args, **kwargs)
File "D:\Python27\lib\site-packages\django\views\decorators\csrf.py" in wrapped_view
  57.         return view_func(*args, **kwargs)
File "D:\Python27\lib\site-packages\rest_framework\views.py" in dispatch
  400.             response = self.handle_exception(exc)
File "D:\Python27\lib\site-packages\rest_framework\views.py" in dispatch
  397.             response = handler(request, *args, **kwargs)
File "D:\Tidal\API\views.py" in put
  343.             return Response(serializer.data, status=status.HTTP_201_CREATED)
File "D:\Python27\lib\site-packages\rest_framework\serializers.py" in data
  573.                 self._data = [self.to_native(item) for item in obj]
File "D:\Python27\lib\site-packages\rest_framework\serializers.py" in to_native
  351.             value = field.field_to_native(obj, field_name)
File "D:\Python27\lib\site-packages\rest_framework\fields.py" in field_to_native
  1035.         value = getattr(self.parent, self.method_name)(obj)
File "D:\Tidal\API\serializers.py" in get_owner_id
  163.         obj.owner_id = Owner.objects.filter(owner_name=obj.owner_name).values_list('owner_id')[0]

Exception Type: AttributeError at /deploy/variable/
Exception Value: 'Varmst' object has no attribute 'owner_name'

我知道'Varmst'没有名为'owner_name'的表。仅仅因为序列化器具有它并不意味着我将使用它....

Answer 1

你可以让你的方法transform_varmst_id更加pythonic：

def transform_varmst_id(self, obj, value):
    maxid = Varmst.objects.latest('varmst_id').varmst_id
    if Varmst.objects.filter(varmst_name=obj.varmst_name).exists():
        obj.varmst_id = Varmst.objects.filter(varmst_name=obj.varmst_name).values_list('varmst_id')[0]
    else:
        obj.varmst_id = maxid + 1
    return obj.varmst_id

在这种特殊情况下，不需要多个return语句。

编辑：

每次调用方法时都不需要创建变量maxid。把它放在条件中：

def transform_varmst_id(self, obj, value):
    if Varmst.objects.filter(varmst_name=obj.varmst_name).exists():
        obj.varmst_id = Varmst.objects.filter(varmst_name=obj.varmst_name).values_list('varmst_id')[0]
    else:
        maxid = Varmst.objects.latest('varmst_id').varmst_id
        obj.varmst_id = maxid + 1
    return obj.varmst_id

这更像是pythonic。您当然可以将else块减少为：

obj.varmst_id = Varmst.objects.latest('varmst_id').varmst_id + 1

但这是品味问题。我更喜欢较短的陈述，纵向优于横向。

django rest framework serializer完全排除了字段

1 个答案: