如何在函数内部以Python形式打印函数名称

Question

def applejuice(q): 
    print THE FUNCTION NAME! 

它应该导致“applejuice”作为字符串。

Answer 1

这也适用于：

import sys 

def applejuice(q): 
    func_name = sys._getframe().f_code.co_name 
    print func_name 

Answer 2

你需要解释你的问题是什么。因为回答你的问题是：

print "applejuice" 

Answer 3

def applejuice(**args): 
    print "Running the function 'applejuice'" 
    pass 

或使用：

myfunc.__name__ 

>>> print applejuice.__name__ 
'applejuice' 

而且，看到how-to-get-the-function-name-as-string-in-python

Answer 4

import traceback 

def applejuice(q): 
    stack = traceback.extract_stack() 
    (filename, line, procname, text) = stack[-1] 
    print procname 

我想这是用于调试，所以你可能想看看traceback module提供的其他程序。他们会告诉你打印整个调用堆栈，异常跟踪等

Answer 5

另一种方式

import inspect 
def applejuice(q): 
    print inspect.getframeinfo(inspect.currentframe())[2] 

Answer 6

def foo(): 
    # a func can just make a call to itself and fetch the name 
    funcName = foo.__name__ 
    # print it 
    print 'Internal: {0}'.format(funcName) 
    # return it 
    return funcName 

# you can fetch the name externally 
fooName = foo.__name__ 
print 'The name of {0} as fetched: {0}'.format(fooName) 

# print what name foo returned in this example 
whatIsTheName = foo() 
print 'The name foo returned is: {0}'.format(whatIsTheName) 

Answer 7

这个网站给了我一个体面的解释sys._getframe.f_code.co_name如何工作，返回函数名称。

http://code.activestate.com/recipes/66062-determining-current-function-name/