从traceback获取最后一个函数的参数？

Question

我从traceback读取Get last函数的调用参数？ 但它不足以回答我的问题。

这真的很困扰我，因为没有调用参数会减慢我的速度，我很确定可以从Python获得这种信息。

这是一个说明问题的例子：

# -*- coding: utf-8 -*-
import sys
import traceback
import inspect
import logging as log

def fl(x):
    # exception is happening here
    y = 5/x
    return y


def fm(x):
    return fl(x-3)


def fn(a, b, c=None):
    return fm(c)


def main():

    try:
        print fn(1, 2, c=3)
    except Exception as e:
        log.error('Unexpected problem.')
        log.error(e)
        traceback.print_exc()
        ### what I need to see is are the call arguments of the last / deepest call: ###
        ### def fl(x) was called with arguments: [(x, 3)]                            ###
        # this does not cut it:
        tb = sys.exc_info()[2]
        traceback.print_tb(tb)
        # this is broken:
        #frames = inspect.getinnerframes(tb)
        #log.error('Argvalues: %s', inspect.getargvalues(frames))
        # not sure:
        frames = inspect.trace()
        argvalues = inspect.getargvalues(frames[0][0])
        log.error('Argvalues: %s', inspect.formatargvalues(*argvalues))



if __name__ == '__main__':
    main()

所以我得到了细节，但调用参数不包含在内：

ERROR:root:Unexpected problem.
ERROR:root:integer division or modulo by zero
Traceback (most recent call last):
  File "sample.py", line 24, in main
    print fn(1, 2, c=3)
  File "sample.py", line 18, in fn
    return fm(c)
  File "sample.py", line 14, in fm
    return fl(x-3)
  File "sample.py", line 9, in fl
    y = 5/x
ZeroDivisionError: integer division or modulo by zero
  File "sample.py", line 24, in main
    print fn(1, 2, c=3)
  File "sample.py", line 18, in fn
    return fm(c)
  File "sample.py", line 14, in fm
    return fl(x-3)
  File "sample.py", line 9, in fl
    y = 5/x
ERROR:root:Argvalues: ()

Answer 1

frames[0][0]表示main功能。 main被称为不带参数，这就是为什么你得到空元组。 将其更改为frames[-1][0]以获取最后一帧。