python – 为什么函数参数之外的“self”会给出“未定义”的错误？

2019年8月4日 222次阅读

看看这段代码：

class MyClass():

    # Why does this give me "NameError: name 'self' is not defined":
    mySelf = self

    # But this does not?
    def myFunction(self):
        mySelf2 = self

基本上我想要一个类来引用自己而不需要专门命名自己的方法,因此我希望自己为类工作,而不仅仅是方法/函数.我怎样才能做到这一点？

编辑：这一点是我试图从类内部引用类名称,类似self.class._name_,这样类名称不会在类的代码中的任何地方进行硬编码,因此它更容易重用代码.

编辑2：从我从下面的答案中学到的,我想要做的是不可能的.我必须找到一种不同的方式.任务放弃了.

编辑3：这是我正在尝试做的事情：

class simpleObject(object):
    def __init__(self, request):
        self.request = request

@view_defaults(renderer='string')
class Test(simpleObject):

    # this line throws an error because of self
    myClassName = self.__class__.__name__

    @view_config(route_name=myClassName)
    def activateTheView(self):
        db = self.request.db
        foo = 'bar'

        return foo

最佳答案 请注意,当您希望类引用自身以使赋值工作时,不会定义self.这是因为(除了被任意命名),self指的是实例而不是类.在可疑的代码行试图运行的时候,还没有类可以引用它.如果有的话,它并不是指代课程.

在方法中,您始终可以使用类型(self).这将获得创建当前实例的MyClass的子类.如果要对MyClass进行硬编码,则该名称将在方法的全局范围内可用.这将允许您执行示例在实际工作时允许的所有内容.例如,您可以在方法中执行MyClass.some_attribute.

您可能希望在创建类之后修改类属性.这可以通过装饰器或临时基础来完成.元类可能更适合.虽然不知道你真正想做什么,但这是不可能的.

更新：

这里有一些代码可以做你想要的.它使用元类AutoViewConfigMeta和一个新的装饰器来标记您希望view_config应用于的方法.我欺骗了view_config装饰器.它会在调用时打印出类名,以证明它可以访问它.元类__new__只循环遍历类字典并查找由auto_view_config装饰器标记的方法.它清除标记并将view_config装饰器应用于适当的类名.

这是代码.

# This just spoofs the view_config decorator.
def view_config(route=''):
    def dec(f):
        def wrapper(*args, **kwargs):
            print "route={0}".format(route)
            return f(*args, **kwargs)
        return wrapper
    return dec

# Apply this decorator to methods for which you want to call view_config with 
# the class name. It will tag them. The metaclass will apply view_config once it 
# has the class name. 
def auto_view_config(f):
    f.auto_view_config = True
    return f

class AutoViewConfigMeta(type):
    def __new__(mcls, name, bases, dict_):
        #This is called during class creation. _dict is the namespace of the class and
        # name is it's name. So the idea is to pull out the methods that need
        # view_config applied to them and manually apply them with the class name.
        # We'll recognize them because they will have the auto_view_config attribute
        # set on them by the `auto_view_config` decorator. Then use type to create
        # the class and return it.

        for item in dict_:
            if hasattr(dict_[item], 'auto_view_config'):  
                method = dict_[item]
                del method.auto_view_config # Clean up after ourselves.
                # The next line is the manual form of applying a decorator.
                dict_[item] = view_config(route=name)(method)  

        # Call out to type to actually create the class with the modified dict.
        return type.__new__(mcls, name, bases, dict_)


class simpleObject(object):
    __metaclass__ = AutoViewConfigMeta 


class Test(simpleObject):

    @auto_view_config
    def activateTheView(self):
        foo = 'bar'

        print foo

if __name__=='__main__':
    t = Test()
    t.activateTheView()

如果您有任何疑问,请告诉我.

点赞