Question

I made a simple script which finds the Square root of a number. The user inputs a number and it finds the square root and shows the result. I want it to check whether the input was a number or not. If it was a number it’ll continue else it’ll show a message, and reset.

I tried using:

while num != int(x):
    print "That is not a valid number"
    return self.page()

But that only shows an error.
Can someone help me out on this?

Here is the code:

import math
import sys

class SqRoot(object):
    def __init__(self):
        super(SqRoot, self).__init__()

        self.page()

    def page(self):
        z = 'Enter a number to find its square root: '
        num = int(raw_input(z))
        sqroot = math.sqrt(num)
        print 'The square root of \'%s\' is \'%s\'' % (num, sqroot)
        choose = raw_input('To use again press Y, to quit Press N: ')
        if choose == 'Y' or choose == 'y':
            return self.page()
        elif choose == 'N' or choose == 'n':
            sys.exit(0)

print "SqRoot Finder v1.0"
print "Copyright(c) 2013 - Ahnaf Tahmid"
print "For non-commercial uses only."
print "--------------------------------"

def main():
    app = SqRoot()
    app()

if __name__ == '__main__':
    main()

Answer 1

One of the python principles is EAFP:

Easier to ask for forgiveness than permission. This common Python coding style assumes the existence of valid keys or attributes and catches exceptions if the assumption proves false.

x = raw_input('Number?')
try:
    x = float(x)
except ValueError:
    print "This doesn't look like a number!"