我看过Get last function's call arguments from traceback?,但它不够具体,无法回答我的问题。
这真的让我感到困扰,因为没有调用参数会减慢我的速度,我很确定可以从Python中获取这种信息。
以下是一个说明问题的示例:
# -*- 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()
所以我得到了详细信息,但调用参数未包含在内:
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: ()
答案 0 :(得分:1)
frames[0][0]
代表main
功能。调用main
时不带参数,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。