Python Code To Ignore Errors

I have a code that stops running each time there is an error. Is there a way to add a code to the script which will ignore all errors and keep running the script until completion?

Solution 1:

You should always 'try' to open files. This way you can manage exceptions, if the file does not exist for example. Take a loot at Python Tutorial Exeption Handling

import sys

try:
    f = open('myfile.txt')
    s = f.readline()
    i = int(s.strip())
except IOError as e:
    print "I/O error({0}): {1}".format(e.errno, e.strerror)
except ValueError:
    print "Could not convert data to an integer."
except:
    print "Unexpected error:", sys.exc_info()[0]
    raise

or

for arg in sys.argv[1:]:
    try:
        f = open(arg, 'r')
    except IOError:
        print 'cannot open', arg
    else:
        print arg, 'has', len(f.readlines()), 'lines'
        f.close()

Do not(!) just 'pass' in the exception block. This will(!) make you fall on your face even harder.

Solution 2:

Where ever your error(s) is happening you can wrap it in a try/except block

for i in loop:
    try:
        code goes here...
    except:
        pass