How to make it so an error happens when user inputs number over 100 or less than 0

Question

Hey does anyone know what error handling I would need to use using the Except command? Thanks and this is the code I am trying to put it to.

try:
    #Get average, amount earned
    school_average = input("Enter your average in school:")
    money_earned = input("How much money did you earn before summer: ")

#determine the resulting vacation
if school_average >= 80 and money_earned >= 500:
    print "You get to go to Europe!"
elif school_average >= 80:
    print "You get to go to California."
else:
    print "You do not get to go away."
except NameError:
    print "Please input a number for your average and amount of money earned."
except:
    print "An error has occurred"

score 1 · Answer 1 · answered Oct 21 '15 at 22:48

1

You can manually throw an error with raise which will be caught be a general except statement (with no exception type specified).

answered Oct 21 '15 at 22:48

TheoretiCAL

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Actually what would the command be for it? NameError or ValueError? – poop Oct 21 '15 at 22:57

score 1 · Answer 2 · edited May 23 '17 at 12:14

You can implement it something like this:

n = int(input("What is it?\n"))

if 100 > n and n > 0:
    print("you did good")
else:
    raise ValueError('it must be between 0 and 100')

If the answer is above 100 or less than 0, it will raise the ValueError execption as such:

What is it?
101
Traceback (most recent call last):
  File "C:/Users/Leb/Desktop/Python/test2.py", line 6, in <module>
    raise ValueError('it must be between 0 and 100')
ValueError: it must be between 0 and 100

For more detailed information check out this answer: Manually raising (throwing) an exception in Python

How to make it so an error happens when user inputs number over 100 or less than 0

2 Answers2