Getting exception details in Python

Question

I have to open & write to about 10 different files all within the same loop. e.g:

for i in range(0,10):
    try:
        a=5
        file1 = open(\"file1.txt         


        

Answer 1

You can use sys.exc_info to get information about the exception currently being handled, including the exception object itself. An IOError exception contains all of the information you need, including the filename, the errno, and a string describing the error:

import sys

try:
    f1 = open('example1')
    f2 = open('example2')
except IOError:
    type, value, traceback = sys.exc_info()
    print('Error opening %s: %s' % (value.filename, value.strerror))

Execution in the try block will obviously still halt after the first exception.