Python try/catch if unable to open file

Question

Below is a simple program of reading a file and reading how many 8s there are in the file:

import sys

if len(sys.argv) == 1:
  print "ERROR: Please put a filename after your Python file!"

else:
  myFile = sys.argv[1] 
  new = open(myFile)
  numEight = 0
  text = new.read()
  line = len(text.splitlines())



for char in text:
  if char == "8":
    numEight = numEight + 1

print "There are ", numEight, "in this text file"

Question is, how do you create a try catch if the filename entered is not correct? or, is there a way to do that in an elif statement?

Possible duplicate of [How do I check whether a file exists using Python?](https://stackoverflow.com/questions/82831/how-do-i-check-whether-a-file-exists-using-python) — OneCricketeer, Sep 19 '17 at 01:11

Henry · Accepted Answer · 2017-09-19T01:18:03.380

0

In a try/except statement, the code that may fail goes in the try part meanwhile the code that should run if an exception occurs goes in the except part.

try:
    myfile = open(filename)
except IOError as E:
    print 'File Not Found'

edited Sep 19 '17 at 01:18

answered Sep 19 '17 at 01:09

Henry

362
2
8

@cricket_007 You are correct, my apologies. I'm getting languages mixed up, I have edited my answer. – Henry Sep 19 '17 at 01:13
@cricket_007 Corrected. – Henry Sep 19 '17 at 01:15
Sure, but docs say "If the file cannot be opened, IOError is raised" :) – OneCricketeer Sep 19 '17 at 01:16
@cricket_007 I was getting mixed up with Python 3. – Henry Sep 19 '17 at 01:18
Python 3 says OSError – OneCricketeer Sep 19 '17 at 01:19
@cricket_007 The original question asked for an exception if there was a mistake in the filename so that it could not be opened, in Python 3 this exception is FileNotFoundError, however a different exception can throw OSError instead. :) – Henry Sep 19 '17 at 01:24

Python try/catch if unable to open file

1 Answers1