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Python 如何正确指定参数以显示方法';Django admin中的输出?_Python_Django_Django Admin - Fatal编程技术网

Python 如何正确指定参数以显示方法';Django admin中的输出?

Python 如何正确指定参数以显示方法';Django admin中的输出?,python,django,django-admin,Python,Django,Django Admin,我有一个这样的模型: class MyModel(TimeStampedModel): fields = #various fields def get_some_info(self): # returns an output class MyModelAdmin(admin.ModelAdmin): list_display = ['my_method'] def my_method(self): return self.g

我有一个这样的模型:

class MyModel(TimeStampedModel):
    fields = #various fields

    def get_some_info(self):
        # returns an output
class MyModelAdmin(admin.ModelAdmin):
    list_display = ['my_method']

    def my_method(self):
        return self.get_some_info()
...
我的管理模型如下所示:

class MyModel(TimeStampedModel):
    fields = #various fields

    def get_some_info(self):
        # returns an output
class MyModelAdmin(admin.ModelAdmin):
    list_display = ['my_method']

    def my_method(self):
        return self.get_some_info()
...

但是管理员站点显示了这个错误:
my_method()只接受1个参数(给定2个)
。怎么了?

该方法应采用两个参数
self
MyModelAdmin
实例,
obj
MyModel
实例

按以下方式替换my_方法将解决您的问题:

def my_method(self, obj):
    return obj.get_some_info()

或者,您可以直接在
列表显示中添加
“获取一些信息”