CODEWITHSUNDEEP
pythontry-except
Chris William
Chris William

Reputation: 1

execution of try and exception in output

I'm having a problem with a program where it runs the exception along with the try command and I don't know what to do since it prints the conversion and the error message.

def main():
    print('What would you like to convert?')
    print('1. Miles to Kilometers')
    print('2. Fahrenheit to Celcius')
    print('3. Gallons to Liters')
    print('4. Pounds to Kilograms')
    print('5. Inches to Centimeters')
    try:
        option = int(input('Menu Option:'))
        if option == 1:
            Inptmiles = float(input('Enter number of miles to convert:'))
            milestokm(Inptmiles)
            usercontinue()
        if option == 2:
            InptFrnht = float(input('Enter degrees that you would like to convert:'))
            FrnhttoClcs(InptFrnht)
            usercontinue()
        if option == 3:
            InptGllns = float(input('Enter number of Gallons you would like to convert:'))
            GllnstoLtrs(InptGlls)
            usercontinue()
        if option == 4:
            InptLbs = float(input('Enter number of pounds you would like to convert:'))
            LbstoKgs(InptLbs)
            usercontinue()
        if option == 5:
            InptInches = float(input('Enter number of Inches you would like to convert:'))
            InchestoCm(InptInches)
            usercontinue()
    except:
        print('Sorry there was an error with your input')
def milestokm(miles):
    convkm = miles * 1.6
    print(f'{miles} Miles is equal to {convkm:.2f} Kilometers')
def FrnhttoClcs(Fahrenheit):
    convClcs = (Fahrenheit - 32) * 5/9
    print(f'{Fahrenheit} Degrees Fahrenheit is equal to {convClcs:.2f} Degrees Celcius')
def GllnstoLtrs(Gallons):
    convLtrs = Gallons * 3.9
    print(f'{Gallons} Gallons is equal to {convLtrs:.2f} Liters')
def LbstoKgs(Pounds):
    convKgs = Pounds * 0.45
    print(f'{Pounds} Pounds is equal to {convKgs:.2f} Kilograms')
def InchestoCm(Inches):
    convCm = Inches * 2.54
    print(f'{Inches} Inches is equal to {convCm:.2f} Centimeters')
    
main()

I tried moving the exception down past the define commands that used the stored variables but it came back as an error.

Upvotes: -3

Views: 87

Answers (2)

AsrtoMichi
AsrtoMichi

Reputation: 36

An error is raised because name 'usercontinue' is not defined. To find the error, I used this snippet of code:

except Exception as e:
    print(e)

Upvotes: 1

Grismar
Grismar

Reputation: 31379

This is exactly why you should never just use try ... except ... without specifying a specific exception. If you re-raise the exception, you'll find that the issue is a missing definition:

except Exception as e:
    print('Sorry there was an error with your input')
    raise e

Output will be something like:

...
Sorry there was an error with your input
Traceback (most recent call last):
  File "<path>\main.py", line 60, in <module>
    main()
  File "<path>\main.py", line 32, in main
    raise(e)
  File "<path>\main.py", line 13, in main
    usercontinue()
    ^^^^^^^^^^^^
NameError: name 'usercontinue' is not defined

So, instead only catch the exception you expect and allow others to pass, after you fix the issue:

except ValueError:
    print('Sorry there was an error with your input')

...

def usercontinue():
    # whatever you planned here
    pass

Upvotes: 1

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