Properly terminate script without error message in Spyder

Question

SO for my project in python, I am taking two inputs say a & b as integer values. Now the code goes like:

import sys
a = input("enter a")
b = input("enter b")
if a < b:
    print(" enter a greater than b and try again")
    sys.exit()

# Rest of the code is here

Now this works fine. But creates an extra statement

An exception has occurred, use %tb to see the full traceback.

SystemExit

And I do not want that as the user may think that the code's functioning is not proper. So is there any way that this statement is not shown or any other function which would exit the code without printing anything except the line which I have written?

NOTE I have tried exit() but it continues to execute the code beneath it. Also, I have noticed this related question but the approaches listed there don't work in this case.

EDIT: I am adding some more information. I need to put this exit function into a user-defined function so that every time the user enters some wrong data, the code will call this user-defined function and will exit the code. If I try to put my code in an if else statement like

def end():
    print("incorrect input try again")
    os.exit()

a = input("enter data")
if a < 10:
    end()
b = input ("enter data")
if b < 20:
   end()
# more code here

I don't know why but I can't even define this user-defined function in the end as it raises the error of undefined function end(). I'm using Python with Spyder on Windows.

Sorry I thought I had solved this problem but unfortunately I hadn't. So please help!! — Ambika Saxena, Jul 14 '17 at 05:48
@moooeeeep currently I am using Spyder for coding on windows and I don' t have any such thing in my code. — Ambika Saxena, Jul 14 '17 at 07:24
Then this seems relevant: https://stackoverflow.com/q/25928377/1025391 (Though maybe a bit outdated as the behavior is slightly different apparantly) You'd better state such information directly in your question. — moooeeeep, Jul 14 '17 at 07:42

Snow · Accepted Answer · 2017-07-14T07:13:55.710

1

You can use os._exit()

a = int(input("enter a"))
b = int(input("enter b"))
if a < b:
    print(" enter a greater than b and try again")
    os._exit(0)

edited Jul 14 '17 at 07:13

answered Jul 13 '17 at 07:17

Snow

1,058
2
19
47

there is no attribute exit() in os. – Ambika Saxena Jul 14 '17 at 07:11
1

Thanks it worked fine without any warnings even!! :) – Ambika Saxena Jul 14 '17 at 07:26

score 1 · Answer 2 · answered Jul 14 '17 at 06:13

1

This seems to work just fine:

import sys

def end():
    print("incorrect input try again")
    sys.exit(1)

a = input("enter data")
if int(a) < 10:
    end()
b = input ("enter data")
if int(b) < 20:
    end()

I used the sys.exit and fixed your conditions to avoid comparing string with integers. I cannot see any additional messages in the output. Only this:

>py -3 test2.py
enter data1
incorrect input try again
>

Please also take notice of this qoute from python docs:

The standard way to exit is sys.exit(n). [os]_exit() should normally only be used in the child process after a fork().

It also works in the repl

answered Jul 14 '17 at 06:13

K. Kirsz

1,384
10
11

If the code is run for the first time, it shows `C:\Users\abc\Anaconda\lib\site-packages\IPython\core\interactiveshell.py:2889: UserWarning: To exit: use 'exit', 'quit', or Ctrl-D. warn("To exit: use 'exit', 'quit', or Ctrl-D.", stacklevel=1) An exception has occurred, use %tb to see the full traceback. SystemExit` and if on the same console if I execute the code again, it does not show the warning but shows this: `An exception has occurred, use %tb to see the full traceback. SystemExit`. I don't want to display these messages. – Ambika Saxena Jul 14 '17 at 07:08
The code exits properly but gives extra statements which I do not need to be printed. – Ambika Saxena Jul 14 '17 at 07:09

Properly terminate script without error message in Spyder

2 Answers2