I have this try/except code:
document = raw_input ('Your document name is ')
try: with open(document, 'r') as a: for element in a: print element
except: print document, 'does not exist'How do I exit the program after I print "[filename] does not exist"? break and pass obviously don't work, and I don't want to have any crashing errors, so sys.exit is not an option.
Please ignore the try part - it's just a dummy.
6 Answers
Use the sys.exit:
import sys
try: # do something
except Exception, e: print >> sys.stderr, "does not exist" print >> sys.stderr, "Exception: %s" % str(e) sys.exit(1)A good practice is to print the Exception that occured so you can debug afterwards.
You can also print the stacktrace with the traceback module.
Note that the int you return in sys.exit will be the return code of your program. To see what exit code your program returned (which will give you information about what happens and can be automated), you can do:
echo $?Using
sys.exit(1)is not a crashing error, it's a perfectly normal way to exit a program. The exit code of 1 is a convention that means something went wrong (you would return 0 in the case of a successful run).
2Just re-raise it. It's more friendly for developer
document = raw_input ('Your document name is ')
try: with open(document, 'r') as a: for element in a: print element
except: print document, 'does not exist' raiseCheck python document in the Raising Exceptions section about re-raise error in except.
You can also put your code in a function and issue a return. You may call it main which you can call from your script.
def main(): document = raw_input ('Your document name is ') try: with open(document, 'r') as a: for element in a: print element except: print document, 'does not exist' return
if __name__ == "__main__": main()In case that you are using an if statement inside a try, you are going to need more than one sys.exit() to actually exit the program.
For example, you are parsing an argument when calling the execution of some file, e.g. $./do_instructions.py 821 such as:
import sys
# index number 1 is used to pass a set of instructions to parse
# allowed values are integer numbers from 1 to 4, maximum number of instructions is 3
arg_vector = "821" # <- pretending to be an example of sys.argv[1]
if len(arg_vector) > 3: sys.exit(2) # <- this will take you out, but the following needs an extra step.
# for an invalid input (8).
for i in arg_vector: # to validate that only numbers are passed as args. try: int(i) # <- 8 is valid so far # value (8) is not valid, since is greater than 4 if (int(i) == 0) or (int(i) > 4): print("Values must be 1-4") # the following call does not takes you out from the program, # but rise the SystemExit exception. sys.exit(2) except SystemExit: # <- needed to catch the previous as the first evaluation # The following parameter "2" is just for this example sys.exit(2) # <- needed to actually interrupt the execution of the program/script. # if there is no "except SystemExit:", the following will be executed when the # previous "if" statement evaluates to True and the sys.exit(2) is called. # # and the "print("Only num...") function will be called, even when the intention # of it is to advice the *user* to use only numbers, since 8 is a number this # shouldn't be executed. except: print("Only numbers are allowed.") sys.exit(2)otherwise, you want to use one try-except block for each evaluation.
Probably not the best practice, but it worked for me:
import sys
close = False
try: if SomethingBadHappend: close = True
except: pass
if close: sys.exit(1)Close dose does not seem to work inside 'try'.
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