在Python中获取调用函数模块的__name__ [英] Get __name__ of calling function's module in Python

问题描述

假设myapp/foo.py包含:

def info(msg):
    caller_name = ????
    print '[%s] %s' % (caller_name, msg)

myapp/bar.py包含:

import foo
foo.info('Hello') # => [myapp.bar] Hello

在这种情况下，我希望将caller_name设置为调用函数模块的__name__属性(即"myapp.foo").该怎么办?

I want caller_name to be set to the __name__ attribute of the calling functions' module (which is 'myapp.foo') in this case. How can this be done?

推荐答案

检出检查模块:

inspect.stack()将返回堆栈信息.

在函数内部，inspect.stack()[1]将返回调用者的堆栈.从那里，您可以获取有关调用者的函数名称，模块等的更多信息.

Inside a function, inspect.stack()[1] will return your caller's stack. From there, you can get more information about the caller's function name, module, etc.

有关详细信息，请参阅文档:

See the docs for details:

http://docs.python.org/library/inspect.html

此外，道格·海尔曼(Doug Hellmann)在其PyMOTW系列中对检查模块进行了很好的撰写:

Also, Doug Hellmann has a nice writeup of the inspect module in his PyMOTW series:

http://pymotw.com/2/inspect/index.html #module-inspect

我认为这是一些您想要执行的代码:

Here's some code which does what you want, I think:

import inspect 

def info(msg):
    frm = inspect.stack()[1]
    mod = inspect.getmodule(frm[0])
    print '[%s] %s' % (mod.__name__, msg)

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