如何在python中另一个类的函数中获取调用者类名? [英] How to get the caller class name inside a function of another class in python?

问题描述

我的目标是模拟一个应用程序的序列图，我需要在运行时有关调用者和被调用者类名的信息.我可以成功检索调用者函数但无法获取调用者类名?

My objective is to stimulate a sequence diagram of an application for this I need the information about a caller and callee class names at runtime. I can successfully retrieve the caller function but not able to get a caller class name?

#Scenario caller.py:

import inspect

class A:

    def Apple(self):
        print "Hello"
        b=B()
        b.Bad()



class B:

    def Bad(self):
        print"dude"
        print inspect.stack()


a=A()
a.Apple()

当我打印堆栈时，没有关于调用者类的信息.那么是否可以在运行时检索调用者类?

When I printed the stack there was no information about the caller class. So is it possible to retrieve the caller class during runtime ?

推荐答案

好吧，在对提示进行一些挖掘之后，我得到了:

Well, after some digging at the prompt, here's what I get:

stack = inspect.stack()
the_class = stack[1][0].f_locals["self"].__class__.__name__
the_method = stack[1][0].f_code.co_name

print("I was called by {}.{}()".format(the_class, the_method))
# => I was called by A.a()

调用时:

➤ python test.py
A.a()
B.b()
  I was called by A.a()

给定文件 test.py:

import inspect

class A:
  def a(self):
    print("A.a()")
    B().b()

class B:
  def b(self):
    print("B.b()")
    stack = inspect.stack()
    the_class = stack[1][0].f_locals["self"].__class__.__name__
    the_method = stack[1][0].f_code.co_name
    print("  I was called by {}.{}()".format(the_class, the_method))

A().a()

不确定从对象以外的东西调用时它会如何表现.

Not sure how it will behave when called from something other than an object.

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