If statement to take input and accept integer

Question

I need to write an if statement that will take user input and accept integers.

Here is the code, but it is only taking the specified integers.

next = raw_input("> ")
if "0" in next or "1" in next:
    how_much = int(next)

You've described a situation and posted some code. What question do you want to ask us? — user2357112, Aug 21 '18 at 22:06
@emsimpson92: There is no `IsNumeric` method, and `isnumeric` means the wrong thing. — user2357112, Aug 21 '18 at 22:07
The key is to `try:` to do that `how_much = int(next)`, `except ValueError:` do whatever you wanted to do in that case. — abarnert, Aug 21 '18 at 22:50

score 0 · Answer 1 · answered Aug 22 '18 at 00:15

There are a couple approaches to this question since I'm not to sure what exactly your looking for. Here they are:

If you want them to be not an integer and you aren't interested in converting the input to an integer check it out below I used the isnumeric() function as stated above:

Python 3 answer below:

enter_num=input('Please Enter a number: \n') # I am prompting the user #for input and the \n means new line so the user would enter a value on #the new line
if enter_num.isnumeric(): #if the value the user entered is a number #proceed into the if statement
    print('You must have entered an integer value') # print this #statement out if the if statement conditon is true

I'm not sure if you are familiar with try blocks so I won't get into it but here is another approach(im also not sure if you are familiar with the isinstance() function):

number=int(input('Please Enter a number: \n')) #prompting the user for #a number

if isinstance(number,int): # if the vlaue is a integer enter the if #statement
    print('this is an integer') #enters the if statement if its true

I'm sorry if my solution helped. But there is also this question which might help: How do I check if raw input is integer in python 2.7?

If statement to take input and accept integer

1 Answers1