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up vote 2 down vote favorite

In a function in Django, the user can send me a number or a string, and I want to know if I received a number or a String (Tip: The number will always be an integer between 1-6)

I want to know if it's possible to detect this and how (with an example), as the number or string I'm getting will tell me what to do next.

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How is the number sent? In a request? –  Martijn Pieters Nov 15 '13 at 14:30
    
Like this: def make_choice(request, option): –  Sascuash Nov 15 '13 at 14:33
    
Then option is always a string. –  Martijn Pieters Nov 15 '13 at 14:34
    
So, trying to convert it into an integer like int(option) will only work if it's a number originally. Am I wrong? –  Sascuash Nov 15 '13 at 14:38

3 Answers 3

up vote 3 down vote accepted

You can try to convert the string to a number using int(), catching the exception:

def isNum(data):
    try:
        int(data)
        return True
    except ValueError:
        return False

This returns True only if the string can be converted to an integer number.

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3  
@Christian: That is what exceptions are for. Not just to detect coding errors. And no, this is not slow, this is actually faster than the alternatives, which is to use a regular expression or a series of string method tests. –  Martijn Pieters Nov 15 '13 at 14:33
    
@Christian: Can you show me anything other than "It's probably slow"? Because that's not much of an argument; do you have proof? –  Martijn Pieters Nov 15 '13 at 14:44
up vote 1 down vote

What about: if isinstance(data, int):

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up vote 1 down vote

I'm assuming your number will still be encased in a string, ie. "1" or "4" or "6" - if that's the case, then there are several ways to do it; you could use a regex to check whether it is a number or not, or you could knock up a function that would look something like this

def isNumber(your_input_string):
    return len(your_input_string) == 1 and your_input_string in "123456"

Since the number will always be between 1 and 6, it can only be a string of length 1, and it must be contained in the string '123456', since... well, those are the allowed numbers.

EDIT:

As pointed by Martijn Pieters in the comments below, that is a roundabout way of doing it; an easier solution would be just to check whether the string is between '1' or '6'. 

def isNumber(your_input_string):
    return len(your_input_string) == 1 and '1' <= your_input_string <= '6'
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2  
You can simply write that as return len(your_input_string) == 1 and your_input_string in "123456" and there are no true and false in python –  thefourtheye Nov 15 '13 at 14:41
2  
This is rather... round-about. return len(your_input_string) == 1 and '1' <= your_input_string <= '6' would do the same, but faster. –  Martijn Pieters Nov 15 '13 at 14:41
    
- I can't believe I overlooked the fact that I'm basically just returning a boolean expression. That's a bit embarassing. It's been a while since I last programmed anything in python, so I wasn't even sure if my syntax was correct. I didn't even remember that you could do comparisons with <= on strings - that's brilliantly elegant solution. –  Tobias Roland Nov 15 '13 at 14:46
    
You do need to test for the length here; '1' <= '16' <= '6' is True as well, due to lexicographic ordering. –  Martijn Pieters Nov 15 '13 at 15:07

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