How can I fix this simple program?

Question

I have written this very simple program in python:

a=input('Enter the grade:')

if int(a)<5:
    print('D')
elif 5<=int(a)<10:
    print('c')
elif 10<=int(a)<15:
    print('B')
elif 15<=int(a)<=20:
    print('A')
elif 20<int(a):
    print('You idiot !')

else :
    print('Write a  number idiot !')

And the program will work if the user write a number, but if they write a string the program will give this error:

Traceback (most recent call last):
  File "C:\Users\sony\Desktop\Grader.py", line 2, in <module>
    if int(a)<5:
ValueError: invalid literal for int() with base 10: 'h'

How can I change the program, so that the users can write anything they want!

Can you please elaborate on what you are exactly trying to do? What are your inputs and expected outputs? — Rubal, Nov 19 '15 at 19:32
By the way. Calling users idiots would be a reason for me to fire somebody. — michel.iamit, Nov 19 '15 at 19:37
Well i'm not your employee ! and that is an exercise program for beginners like me which my friend decided to make it a little fun ! do you think , we would call users idiot in a real program (like a project which we later have to write at the end of the semester) ? — user82087, Nov 20 '15 at 11:02

Nir Alfasi · Answer 1 · 2015-11-20T02:44:15.617

1

modify:

a=input('Enter the grade:')

to:

a = None
while not a:
    try:
        a = int(input('Enter the grade:'))
    except ValueError:
        print("please enter a valid integer!)
    else:
        break

edited Nov 20 '15 at 02:44

answered Nov 19 '15 at 19:37

Nir Alfasi

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What if they type `0` ? – khelwood Nov 19 '15 at 19:41
@khelwood good catch, here you go :) – Nir Alfasi Nov 19 '15 at 19:47
Why not just `while True`? Also, if you don't use `e`, then no need `as e`. – Remi Guan Nov 20 '15 at 02:17
1

@KevinGuan you can do while True, indeed it saves you one character but doesn't explain what are we waiting for to happen in the while loop. As for `e`, true - I'll modify it. – Nir Alfasi Nov 20 '15 at 02:43

score 0 · Answer 2 · answered Nov 19 '15 at 19:34

0

You can ether check the type of the variable or do a try / except. (Last is the easiest way, and see sufficiënt here)

answered Nov 19 '15 at 19:34

michel.iamit

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score 0 · Answer 3 · answered Nov 19 '15 at 19:36

You can use this function to check whether the input value is an integer or not:

In [1]: def is_int(value):
...:     try:
...:         int(value)
...:         return True
...:     except ValueError:
...:         return False
...:     

In [2]: is_int(6)
Out[2]: True

In [3]: is_int('something')
Out[3]: False

Hope this helps!

How can I fix this simple program?

3 Answers3