python 2.7 converting to 3.4. Error in 3.4, when no error in 2.7

Question

I have tested and used this code in python 2.4. However, as soon as I moved onto python 3.4, an error comes up - "TypeError: unsupported operand type(s) for +: 'int' and 'str'. I would greatly appreciate it if you managed to find the reason why this is happening. I will post the code I used below

alphabetL = 'abcdefghijklmnopqrstuvwxyz'
alphabetC = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
number = '0123456789'
space = ' '
Ll = len(alphabetL)
Lc = len(alphabetC)
Ln = len(number)
Lall = Ll + Lc + Ln
Question1 = input('Hello, please insert the message you want encrypted: ')
key1 = input('Please insert the key you want used [Keep between 1 and 26]: ')
cipher = ''
cipher2 = ''


for A in Question1:
    if A in alphabetL:
        cipher += alphabetL[(alphabetL.index(A)+key1)%Ll]
    elif A in alphabetC:
        cipher += alphabetC[(alphabetC.index(A)+key1)%Lc]
    elif A in number:
        cipher += number[(number.index(A)+key1)%Ln]
    elif A in space:
        cipher += space
    else:
        print ('Error, please use abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789')
print (cipher)

`input` in 3.x is like `raw_input` (which you should *already have been using*) in 2.x. — jonrsharpe, Nov 16 '15 at 12:19
Because you have a string (e.g. `'1'`) not an integer (`1`), which is *exactly what the error message is telling you*. — jonrsharpe, Nov 16 '15 at 12:25
okay, calm down. It's clear I'm a novice coder so could you please help me in a more practical way. — BigTommyVernon, Nov 16 '15 at 12:28
Possible duplicate of [How can I read inputs as integers in Python?](http://stackoverflow.com/questions/20449427/how-can-i-read-inputs-as-integers-in-python) — jonrsharpe, Nov 16 '15 at 12:29

score 0 · Answer 1 · answered May 05 '16 at 14:15

You can't use a string as an index

cipher += alphabetL[(alphabetL.index(A)+key1)%Ll]

alphabetL.index(A) is an integer and key1 is a string
so you must convert key1 to integer.

key1 = int(input('Please insert the key you want used [Keep between 1 and 26]: '))

python 2.7 converting to 3.4. Error in 3.4, when no error in 2.7

1 Answers1