CODEWITHSUNDEEP
pythonpython-2.7inputwhile-loop
Markus Rexhepi-Lindberg
Markus Rexhepi-Lindberg

Reputation: 167

While loop for a raw_input gives strange output in Python

No mather what input I give I get the output "I think you're to old to go on an adventure. Your adventure ends here." Like if I input 17 I get "I think you're to old to go on an adventure. Your adventure ends here." and if I input 18-59 I get "I think you're to old to go on an adventure. Your adventure ends here."

while True:
    age = raw_input("\n So tell me " + str(name) + ", how old are you?: ")
    if age.isdigit() == False:
        print "\n Please, put only in numbers!"
        time.sleep(3)
    elif age < 18:
        print "\n %s. A minor. You got to be atleast 18 to go on this adventure. Your adventure ends here." % age
        time.sleep(7)
        exit(0) 
    elif age >= 60:
        print "\n %s. I think you're to old to go on an adventure. Your adventure ends here." % age
        time.sleep(5)
        exit(0)
    else:
        print "\n %s. You're starting to get old." % age
        break

Upvotes: 0

Views: 139

Answers (2)

TheSoundDefense
TheSoundDefense

Reputation: 6935

The problem is that raw_input always returns a string object, and those don't really compare to int types. You need to do some type conversion.

If you want to use isdigit to test if the input is a number, then you should proceed this way:

while True:
    age = raw_input("\n So tell me " + str(name) + ", how old are you?: ")
    if age.isdigit() == False:
        print "\n Please, put only in numbers!"
        time.sleep(3)
    elif int(age) < 18:
        print "\n %s. A minor. You got to be atleast 18 to go on this adventure. Your adventure ends here." % age
        time.sleep(7)
        exit(0) 
    elif int(age) >= 60:
        print "\n %s. I think you're to old to go on an adventure. Your adventure ends here." % age
        time.sleep(5)
        exit(0)
    else:
        print "\n %s. You're starting to get old." % age
        break

However, you can simplify the code a little by converting to an integer right away, and just catching the exception if it's an invalid string:

while True:
    try:
        age = int(raw_input("\n So tell me " + str(name) + ", how old are you?: "))
        if age < 18:
            print "\n %s. A minor. You got to be atleast 18 to go on this adventure. Your adventure ends here." % age
            time.sleep(7)
            exit(0) 
        elif age >= 60:
            print "\n %s. I think you're to old to go on an adventure. Your adventure ends here." % age
            time.sleep(5)
            exit(0)
        else:
            print "\n %s. You're starting to get old." % age
            break
    except ValueError:
        print "\n Please, put only in numbers!"
        time.sleep(3)

Upvotes: 1

Cory Kramer
Cory Kramer

Reputation: 117856

You need to compare your input as an int

age = int(raw_input("\n So tell me " + str(name) + ", how old are you?: "))

Otherwise you are comparing a str to an int. See the following example

>>> 5 < 10
True
>>> type(5)
<type 'int'>

>>> '5' < 10
False
>>> type('5')
<type 'str'>

In Python 2.x, comparing values of different types generally ignores the values and compares the types instead. Because str >= int, any string is >= any integer. In Python 3.x you get a TypeError instead of silently doing something confusing and hard to debug.

Upvotes: 3

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