Check if input is int or float in range [duplicate]-CodePudding

I want a while loop to store an input in a variable and check if it is an integer or a float before printing a statement and breaking the loop. If the variable fails the check I want it to continue back to the top of the loop and ask for the input again until it gets a valid response.

Somehow regardless of what the input is, the loop asks for an input again and seems to skip the if conditional block.

def FeelingCalc():

    Monday = 0

    while Monday == 0:
        Monday = input("On a scale of 1 to 10, how did you feel on Monday? ")
        if isinstance(Monday,(int,float)) and 11 > float(Monday) > 0:
            print("Alright. Let's move on.")
            break
        else:
            print("Please type a whole number between 1 and 10.")
            Monday = 0
            continue

I have tried different solutions for the if conditional like:

if Monday != type(int)

but that doesn't seem to do anything either. It still goes to the else block.

CodePudding user response：

You always get string when you use input. So, you have to convert Monday to an integer or float.

Monday = int(input("On a scale of 1 to 10, how did you feel on Monday? "))

EDIT

You might want to use try block:

try:
   Monday = float(input("On a scale of 1 to 10, how did you feel on Monday?"))
except: 
   Monday = 0

CodePudding user response：

The problem is here:

if isinstance(Monday,(int,float)) and 11 > float(Monday) > 0:

Since Monday is derived from input() it will always be a string. If it wasn't a string, but an int or float, there would be no need to cast it to float. You probably want something like this:

try:
    x = float(input())
except:
    print("Please enter a number ")

if x > 11 or x < 0:
    print("Number must be in range 1-10")

EDIT if you don't want anything except int (as the other answers assume), replace float with int.