Can you use input to accept both an integer and a string?

Question

I have a small script I have been working on for practice with Python. I am having trouble getting my input to accept a number for an if statement and also accept string as lower case.

I want to tell my script for any input if the user types '99' then close the program. So far it works where I have int(input()), but it won't work where I have input(). What am I doing wrong or is it not something I can do?

Right now my if statement looks like:

if choice1 == 99:
    break

Should I just make 99 into a string by quoting it?

Maybe something like this:

if choice1 == "99":
    break

Here is the script:

global flip
flip = True
global prun
prun = False

def note_finder(word):
    prun = True
    while prun == True:
        print ('\n','\n','Type one of the following keywords: ','\n','\n', keywords,)
        choice2 = input('--> ').lower()
        if choice2 == 'exit':
            print ('Exiting Function')
            prun = False
            start_over(input)
        elif choice2 == 99:   # this is where the scrip doesnt work
            break             # for some reason it skips this elif
        elif choice2 in notes:
            print (notes[choice2],'\n')
        else:
            print ('\n',"Not a Keyword",'\n')

def start_over(word):
    flip = True
    while flip == True:
        print ('# Type one of the following options:\n# 1 \n# Type "99" to exit the program')
        choice1 = int(input('--> '))
        if choice1 == 99:
            break
        elif choice1 < 1 or choice1 > 1:
            print ("Not an option")
        else:
            flip = False
            note_finder(input)


while flip == True:
    print ('# Type one of the following options:\n# 1 \n# Type "99" to exit the program')
    choice1 = int(input('--> '))
    if choice1 == 99:
        break
    elif choice1 < 1 or choice1 > 1:
        print ("Not an option")
    else:
        flip = False
        note_finder(input)

If you don't specify a type then it will default to string so numbers can also be accepted. — Joshua Nixon
– Joshua Nixon, Commented Mar 15, 2017 at 20:57
EDIT: my last comment half answered itself. All input() is a string until you convert it. Re-reading your question, there is nothing fundamentally wrong with your approach, but you're going to need to shore it up with try/except etc. for invalid inputs that can't be converted to an int — roganjosh
– roganjosh, Commented Mar 15, 2017 at 20:58
You could always check what data type it is before the if statement using 'isinstance(var, type)' or 'type(var) == int' — Joshua Nixon
– Joshua Nixon, Commented Mar 15, 2017 at 21:00
I added some notes in the scrip to show where the problem is. For some reason the scrip skips the elif choice2 == 99: statement — Mike - SMT
– Mike - SMT, Commented Mar 15, 2017 at 21:03
In that case, then yes, I'd probably use elif choice2 == '99': purely because your code seems set up to expect "exit" first, and any string will throw an error for elif int(choice2) == 99:. It's perhaps not the cleanest approach — roganjosh
– roganjosh, Commented Mar 15, 2017 at 21:07

James R · Accepted Answer · 2017-03-15 21:19:17Z

2

So input() always returns a string. You can see the docs here:

https://docs.python.org/3/library/functions.html#input

What you could do is something like this:

choice2 = input('--> ')

if choice2.isnumeric() and (int(choice2) == 99):
    break

this avoids you to type check and catch errors that aren't important.

see below for how isnumeric works with different numeric types:

In [12]: a='1'

In [13]: a.isnumeric()
Out[13]: True

In [14]: a='1.0'

In [15]: a.isnumeric()
Out[15]: False

In [16]: a='a'

In [17]: a.isnumeric()
Out[17]: False

edited Mar 15, 2017 at 21:19

answered Mar 15, 2017 at 21:15

James R

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1 Comment

mkrieger1 Over a year ago

You just explained in your answer that the result of input is a string. Why do you suggest do compare it to a number then?

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