如何从 Python 函数或方法中获取函数或方法的名称？
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How do I get the name of a function or method from within a Python function or method?
提问by Daryl Spitzer
I feel like I should know this, but I haven't been able to figure it out...
我觉得我应该知道这一点，但我一直无法弄清楚......
I want to get the name of a method--which happens to be an integration test--from inside it so it can print out some diagnostic text. I can, of course, just hard-code the method's name in the string, but I'd like to make the test a little more DRY if possible.
我想从它内部获取方法的名称——它恰好是一个集成测试——以便它可以打印出一些诊断文本。当然，我可以在字符串中硬编码方法的名称，但如果可能的话，我想让测试更加干燥。
采纳答案by spiv
The answers involving introspection via inspectand the like are reasonable. But there may be another option, depending on your situation:
涉及introspection viainspect等的答案是合理的。但可能还有另一种选择，具体取决于您的情况：
If your integration test is written with the unittestmodule, then you could use self.id()within your TestCase.
如果您的集成测试是使用unittest模块编写的，那么您可以self.id()在您的 TestCase 中使用。
回答by mhawke
This seems to be the simplest way using module inspect:
这似乎是使用 module 的最简单方法inspect：
import inspect
def somefunc(a,b,c):
print "My name is: %s" % inspect.stack()[0][3]
You could generalise this with:
您可以将其概括为：
def funcname():
return inspect.stack()[1][3]
def somefunc(a,b,c):
print "My name is: %s" % funcname()
Credit to Stefaan Lippenswhich was found via google.
感谢斯特凡Lippens这是通过谷歌找到。
回答by nosklo
This decorator makes the name of the method available inside the function by passing it as a keyword argument.
此装饰器通过将其作为关键字参数传递，使该方法的名称在函数内可用。
from functools import wraps
def pass_func_name(func):
"Name of decorated function will be passed as keyword arg _func_name"
@wraps(func)
def _pass_name(*args, **kwds):
kwds['_func_name'] = func.func_name
return func(*args, **kwds)
return _pass_name
You would use it this way:
你会这样使用它：
@pass_func_name
def sum(a, b, _func_name):
print "running function %s" % _func_name
return a + b
print sum(2, 4)
But maybe you'd want to write what you want directly inside the decorator itself. Then the code is an example of a way to get the function name in a decorator. If you give more details about what you want to do in the function, that requires the name, maybe I can suggest something else.
但也许您想直接在装饰器内部编写您想要的内容。然后代码是在装饰器中获取函数名称的方法示例。如果您提供有关您想在函数中执行的操作的更多详细信息，则需要名称，也许我可以提出其他建议。
回答by Kevin Little
# file "foo.py"
import sys
import os
def LINE( back = 0 ):
return sys._getframe( back + 1 ).f_lineno
def FILE( back = 0 ):
return sys._getframe( back + 1 ).f_code.co_filename
def FUNC( back = 0):
return sys._getframe( back + 1 ).f_code.co_name
def WHERE( back = 0 ):
frame = sys._getframe( back + 1 )
return "%s/%s %s()" % ( os.path.basename( frame.f_code.co_filename ),
frame.f_lineno, frame.f_code.co_name )
def testit():
print "Here in %s, file %s, line %s" % ( FUNC(), FILE(), LINE() )
print "WHERE says '%s'" % WHERE()
testit()
Output:
输出：
$ python foo.py
Here in testit, file foo.py, line 17
WHERE says 'foo.py/18 testit()'
Use "back = 1" to find info regarding two levels back down the stack, etc.
使用“back = 1”来查找有关返回堆栈的两个级别的信息等。
