Python - Get the source code of the line that called me

Question

Using the python inspect module, in a function, I would like to get the source code of the line that called that function.

So in the following situation:

def fct1():
    # Retrieve the line that called me and extract 'a'
    return an object containing name='a'

a = fct1()

I would like to retrieve the string "a = fct1()" in fct1

All I can do so far is to retrieve the code of the whole module with :

code = inspect.getsource(sys._getframe().f_back)

Please note that fct1() can be called many times in the main module.

Eventually, what I want is to retrieve the variable name "a" which is easy if I can get s = "a = fct1()" in fct1() :

a_name = s.split("=")[0].strip()

Answer 1

一个非常愚蠢的解决方案是捕获堆栈跟踪并获取第二行：

import traceback

def fct1():
    stack = traceback.extract_stack(limit=2)
    print(traceback.format_list(stack)[0].split('\n')[1].strip())  # prints "a = fct1()"
    return None

a = fct1()

@jtlz2 在装饰器中请求它

import traceback

def add_caller(func):
    def wrapper(*args, **kwargs):
        stack = traceback.extract_stack(limit=2)
        func(*args, caller=traceback.format_list(stack)[0].split('\n')[1].strip(), **kwargs)
    return wrapper

@add_caller
def fct1(caller):
    print(caller)

fct1()

它确实有效