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?

Below is the code:

import sys
import tldextractdef main(argv):in_file = argv[1]f = open(in_file,'r')urlList = f.readlines()f.close()destList = []for i in urlList:print istr0 = ifor ch in ['\n','\r']:if ch in str0:str0 = str0.replace(ch,'')str1 = str(tldextract.extract(str0))str2 = i.replace('\n','') + str1.replace("ExtractResult",":")+'\n'destList.append(str2)f = open('destFile.txt','w')for i in destList:f.write(i)f.close()print "Completed successfully:"if __name__== "__main__":main(sys.argv)

Many thanks

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 systry: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', argelse: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.