Adding Exception Handling to a Python Script

The first expression in the exception_handling.py script reads from sys.argv, which is a list. If the script is run without any additional positional arguments, there won't be an item at the index of 1 and we'll receive an IndexError. Let's add exception handling around this line to catch the error, print an error message about the number of required arguments, and then use sys.exit(1) to stop the script and exit with a non-zero status code.

~/exception_handling.py

import sys

# 1) Fetch name from the first argument passed to the script.
# If there is no argument, then exist with a readable error message.
# The potential error here is an IndexError.
try:
    name = sys.argv[1]
except IndexError:
    print("Error: please provide a name and a repeat count to the script.")
    sys.exit(1)

# Remainder of file omitted

Now if we run the script without any arguments, we should see this error message:

$ python3.7 exception_handling.py
Error: please provide a name and a repeat count to the script.
$ python3.7 exception_handling.py Kevin
Traceback (most recent call last):
  File "exception_handling.py", line 16, in <module>
    repeat_count = int(sys.argv[2])
IndexError: list index out of range

This next error is related to the next bit of logic in the script.

When we set the repeat_count variable, we index sys.argv at the index of 2 so we can once again run into an IndexError. In that case, we'll want to provide the same error message we did for the name variable. Additionally, we can also run into a ValueError if the value provided cannot be cast to an integer, so we'll need to have a second except statement that handles that and provides a more useful error message.

~/exception_handling.py

# previous code omitted

# 2) Convert the second argument to be an integer so that we can repeatedly print out
# the name argument. We should either a ValueError or an IndexError if there weren't enough
# arguments provided.
try:
    repeat_count = int(sys.argv[2])
except IndexError:
    print("Error: please provide a name and a repeat count to the script.")
    sys.exit(1)
except ValueError:
    print("Error: the repeat count needs to be a number.")
    sys.exit(1)

# remaining code omitted

The last section of this script deals with interacting with a file, so there are numerous potential errors we could run into. Depending on the version of Python we use to run the script, we could get either an OSError or a PermissionError if we don't have permissions to interact with a file. To make this script backward-compatible, we'll catch OSError since PermissionError inherits from OSError and would also be caught. Additionally, we're going to catch IOError since this is another common error that happens when interacting with files. These error types will both be caught by the same except statement. Our requirements state that we need to print to the screen if we can't write to a file, and we're going to keep our try block small by writing to the file in the else branch if we can read from it.

~/exception_handling.py

# previous code omitted

# 3) Open a file called `name_repeated.txt` in the `root_files` directory and write a line for each time the name was
# repeated. If the user running the script doesn't have write permissions for the directory that
# we're writing to then we might see a PermissionError. PermissionError is new as of 3.3, so let's instead catch
# OSError and IOError to make the script more backward compatible. If an error is caught print to the screen instead.
try:
    f = open("root_files/name_repeated.txt", "w")
except (OSError, IOError) as err:
    print("Error: unable to write file (", err, ")")
    for i in range(1, repeat_count + 1):
        print(i, "-", name)
else:
    names = [name + "\n" for i in range(1, repeat_count + 1)]
    f.writelines(names)
    f.close()

Now we've successfully handled all the common errors we might run into when running this script.