在Python中的另一个函数中获取调用函数名称？

If you have 2 functions like:

如果你有两个功能，如：

def A
def B

and A calls B, can you get who is calling B inside B, like:

和A打电话给B，你能得到谁在B里面叫B，比如：

def A () :
    B ()

def B () :
    this.caller.name

5 个解决方案

#1

108

You can use the inspect module to get the calling stack. It returns a list of frame records. The third element in each record is the caller name. What you want is this:

您可以使用inspect模块获取调用堆栈。它返回一个帧记录列表。每条记录中的第三个元素是调用者名称。你想要的是这个：

>>> import inspect
>>> def f():
...     print inspect.stack()[1][3]
...
>>> def g():
...     f()
...
>>> g()
g

Of course, it is a good idea to check that enough frame records exist before trying to access a particular index.

当然，在尝试访问特定索引之前检查是否存在足够的帧记录是个好主意。

#2

There are two ways, using sys and inspect modules:

有两种方法，使用sys和inspect模块：

sys._getframe(1).f_code.co_name
sys._getframe（1）.f_code.co_name
inspect.stack()[1][3]
inspect.stack（）[1] [3]

The stack() form is less readable and is implementation dependent since it calls sys._getframe(), see extract from inspect.py:

stack（）形式的可读性较差，并且依赖于实现，因为它调用sys._getframe（），请参阅inspect.py中的extract：

def stack(context=1):
    """Return a list of records for the stack above the caller's frame."""
    return getouterframes(sys._getframe(1), context)

#3

Note (June 2018): today, I would probably use inspect module, see other answers

注意（2018年6月）：今天，我可能会使用检查模块，请参阅其他答案

sys._getframe(1).f_code.co_name like in the example below:

sys._getframe（1）.f_code.co_name，如下例所示：

>>> def foo():
...  global x
...  x = sys._getframe(1)
...
>>> def y(): foo()
...
>>> y()
>>> x.f_code.co_name
'y'
>>>

Important note: as it's obvious from the _getframe method name (hey, it starts with an underscore), it's not an API method one should be thoughtlessly rely on.

重要说明：从_getframe方法名称中可以明显看出（嘿，它以下划线开头），它不是一种应该不加思索地依赖的API方法。

#4

This works for me! :D

这适合我！：d

>>> def a():
...     import sys
...     print sys._getframe(1).f_code.co_name
...
>>> def b():
...     a()
...
...
>>> b()
b
>>>

#5

you can user the logging module and specify the %(filename)s option in BaseConfig()

您可以使用日志记录模块并在BaseConfig（）中指定％（filename）选项

import logging
logging.basicConfig(filename='/tmp/test.log', level=logging.DEBUG, format='%(asctime)s | %(levelname)s | %(funcName)s |%(message)s')

def A():
    logging.info('info')

#1

108