相关疑难解决方法(0)

当我的类被子类化时,有没有办法触发代码？

class SuperClass:
    def triggered_routine(subclass):
        print("was subclassed by " + subclass.__name__)

magically_register_triggered_routine()

print("foo")

class SubClass0(SuperClass):
    pass

print("bar")

class SubClass1(SuperClass):
    print("test")

应输出

foo
was subclassed by SubClass0
bar
test
was subclassed by SubClass1

在Python in a Nutshell(第2版)一书中,有一个示例使用
旧样式类来演示如何以经典分辨率顺序解析方法,
以及它与新顺序的不同之处.

我通过重写新样式的示例尝试了相同的示例,但结果与使用旧样式类获得的结果没有什么不同.我用来运行该示例的python版本是2.5.2.以下是示例:

class Base1(object):  
    def amethod(self): print "Base1"  

class Base2(Base1):  
    pass

class Base3(object):  
    def amethod(self): print "Base3"

class Derived(Base2,Base3):  
    pass

instance = Derived()  
instance.amethod()  
print Derived.__mro__  

调用instance.amethod()打印Base1,但根据我对MRO的理解,新类型的输出应该是Base3.通话Derived.__mro__打印:

(<class '__main__.Derived'>, <class '__main__.Base2'>, <class '__main__.Base1'>, <class '__main__.Base3'>, <type 'object'>)

我不确定我对新样式类的MRO的理解是不正确的还是我做了一个我无法察觉的愚蠢错误.请帮助我更好地了解MRO.

在Python2.7这个代码可以很好地工作,__getattr__在MetaTable 运行.但是在Python 3.5中它不起作用.

class MetaTable(type):
    def __getattr__(cls, key):
        temp = key.split("__")
        name = temp[0]
        alias = None

        if len(temp) > 1:
            alias = temp[1]

        return cls(name, alias)


class Table(object):
    __metaclass__ = MetaTable

    def __init__(self, name, alias=None):
        self._name = name
        self._alias = alias


d = Table
d.student__s

但是在Python 3.5中我得到了一个属性错误:

Traceback (most recent call last):
  File "/Users/wyx/project/python3/sql/dd.py", line 31, in <module>
    d.student__s
AttributeError: type object 'Table' has no attribute 'student__s'