why is my try and except not working correctly? python 3

Question

I don't know why my try and except isn't working properly. I want it to continue anytime the user inputs either the string "y" or "n", but if the user doesn't input any of those then print the error.

output:

userInput = input("Are both players ready to start? (y/n): ")
k (userInput)
Here we go!

expected output:

userInput = input("Are both players ready to start? (y/n): ")
k (userInput)
Wrong Input, Try Again

try:
  if userInput == "y":
    print("Here we go! ")
    print()
  elif userInput == "n":
    print("Too bad we're starting anyways")
except:
  print("Wrong Input Try Again)

There is no exception to be thrown if the user adds no input, just neither of your conditions are True. Then there's no outer loop to make it ask again. — roganjosh, Dec 04 '18 at 21:48
There's a missing string quote, it should read `print("Wrong Input Try Again")`.That's probably not your answer though — Seraf, Dec 04 '18 at 21:48

Axel Puig · Answer 1 · 2018-12-05T12:58:49.090

2

There is no error raised in your code, so except is never called. If you want to raise your own exception, use the raise keyword. Details here : https://docs.python.org/2/tutorial/errors.html#raising-exceptions

edited Dec 05 '18 at 12:58

answered Dec 04 '18 at 21:49

Axel Puig

1,304
9
19

Seraf · Accepted Answer · 2018-12-04T22:14:07.173

1

In the try block, you are not raising an error so there is nothing to catch. Adding a raise should do the trick.

userInput = input("Are both players ready to start? (y/n): ")
try:
  if userInput == "y":
    print("Here we go! ")
    print()
  elif userInput == "n":
    print("Too bad we're starting anyways")
  else:
    raise ValueError("What's up with that?")
except:
  print("Wrong Input Try Again")

As suggested by @ctrl-alt-delor you can also skip the try/except block and only use an if/else block. This snippet should do the same:

  if userInput == "y":
    print("Here we go! ")
    print()
  elif userInput == "n":
    print("Too bad we're starting anyways")
  else:
    print("Wrong Input Try Again")

edited Dec 04 '18 at 22:14

answered Dec 04 '18 at 21:53

Seraf

850
1
17
34

it gave me the error: raw_input is not defined. I think it's because this is python 3.7 – AnonymousC_Lo Dec 04 '18 at 22:06
Ok nevermind that part, I just realized I'm on python 2.7. I will edit my answer – Seraf Dec 04 '18 at 22:07
raw_input of python 2, is now input in python 3 – ctrl-alt-delor Dec 04 '18 at 22:07
While your code will work. Why throw and catch an exception, when you can just deal with the input error in the else? – ctrl-alt-delor Dec 04 '18 at 22:09
so it'll say "Wrong Input Try again" but it will still continue my code – AnonymousC_Lo Dec 04 '18 at 22:09
Maybe he wants to handle input interruptions? @ctrl-alt-delor I'm not sure. – Seraf Dec 04 '18 at 22:10
1

@ChamongLo `if/elif/else` will do the same! – Seraf Dec 04 '18 at 22:11
ok so i have written the game of war card game, and i ask the user if they are ready to start, but if they input something other than the strings y and n, I want it to say "Wrong input try again", until they put in the write input – AnonymousC_Lo Dec 04 '18 at 22:12
Maybe, so will this code handle input interruptions? – ctrl-alt-delor Dec 04 '18 at 22:12
error and exceptions – AnonymousC_Lo Dec 04 '18 at 22:14
I add the suggestion from @ctrl-alt-delor in the answer, maybe you can try if that works? It will make your code easier to read in my opinion – Seraf Dec 04 '18 at 22:15
okay it think i found my problem. I had a function after my statements and was continuing because the input was after them. Will i need a while loop to keep asking until the question is answered properly? – AnonymousC_Lo Dec 04 '18 at 22:19

why is my try and except not working correctly? python 3

2 Answers2