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.
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.
What about: if isinstance(data, int):
isinstance('7', int) will return FalseI'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'
return len(your_input_string) == 1 and your_input_string in "123456" and there are no true and false in pythonreturn len(your_input_string) == 1 and '1' <= your_input_string <= '6' would do the same, but faster.
'1' <= '16' <= '6' is True as well, due to lexicographic ordering.
optionis always a string.