Failing While loop in Python

You are on the safer side with raw_input in Python 2, because it will always return a string, whereas input will try to evaluate the given string, which is dangerous.

def askFor(prompt):
    """ ask for input, continues to ask until an integer is given
    """
    number = None
    while True:
        try:
            number = int(raw_input(prompt))
            break
        except ValueError:
            pass # or display some hint ...
    print(number) # or return or what you need ...

The problem is that input() returns the raw user input evaluated as a python expression. So when you input something like "someinput" on the console here is what happens:

1. The raw input is captured from the console as a string
2. That string is then parsed as a python expression (so imagine you simply type this string into your program, in this case, it becomes a variable with the name "someinput"
3. When isinstance() is called on this python expression, the python interpreter can't find a variable named "someinput" which is defined - so it throws an exception.

What you really want to do is use raw_input(), this will return a string representing the user's input. Since you are receiving a string from the console now though, you need some way of checking if a string is an int (because isinstance will always return a string)

Now to determine if the string is a number, you could use something like isdigit() - but be careful - this function will validate any string that contains only digits (so 0393 would be valid).

Here are the corrections that need to be made:

def askFor(request):
   """Program asks for input, continues to ask until an integer is given"""
   num = ' '
    while (num.isdigit() == False):
        num = raw_input(request)
    else:
        return num

This is because input() is failing on non-integer input (this has to do with the fact that input() is calling eval() on the string your inputting (see python builtins).

raw_input() does not. (in fact, input() calls raw_input())

Try this instead:

while True:
    num = raw_input(request)
    if num.isdigit():
        break

The isdigit() function checks to see if each character in a string is a digit ('0' - '9')

input evaluates whatever's typed in as Python code, so if the user types something non-sensical in, it'll produce an error. Or worse, it can run arbitrary commands on your computer. Usually, you want to use raw_input and convert the string value it returns safely yourself. Here, you just need to call int on the string, which can still produce an error if the user enters a non-int, but we can handle that by looping.

And voila:

def read_int(request):
    while True:
        try:
            return int(raw_input(request))
        except ValueError:
            pass

Note, there's no need for break or else; just return the value once you have parsed it successfully!

Well first off, your indentation if a bit off (the while loop's indentation level does not match the previous level).

def askFor(request):
    """Program asks for input, continues to ask until an integer is given"""
    num, isNum = '', False # multiple assignment
    while not isNum:
        num = input(request)
        if type(num) in (int, long): # if it is any of these
           isNum = True
           break  
        else: isNum = False
    return num # at the end of the loop, it will return the number, no need for an else

Another way to do it:

def askFor(request):
    """Program asks for input, continues to ask until an integer is given"""
    num, isNum = '', False
    while not isNum:
        num = raw_input(request)
        try: # try to convert it to an integer, unless there is an error
            num = int(num)
            isNum = True
            break
        except ValueError:
            isNum = False
    return num