'How to format a user input into a different out put?Python

How can from an input (10 number): XXXXXXXXXX

convert into : X.XX.XX.XX.XXXX

So far this is my code:

creat_number_account = input("Account Number:")

def c_n_account(creat_number_account):
    
    if len(creat_number_account) <10:
        print("This account require 10 numbers")
        
    elif n_list[10] == "0":
            print("The last number can't be: '0'")
    else:
        list =[]
        for n in creat_number_account:
            list+= n
        print(f'{list [0]}.{list[1][2]}.'
    f'{list[3][4]}.{list[5][6]}.{list[7][8][9][-1]}')

print(c_n_account(creat_number_account))

OUTPUT:

print(f'{list [0]}.{list[1][2]}.'
IndexError: string index out of range


Solution 1:[1]

Adding to answer by @Sefan, keep prompting the user until the correct format is recevied:

n = input("Account Number:")

def c_n_account(n):
    while True:
        if len(n) < 10 or len(n) > 10:
            print("This account requires 10 numbers only")
            n = input("Account Number:")
            continue            
        elif n[9] == "0":
             print("The last number can't be: '0'")
             n = input("Account Number:")
             continue    
        else:
            act = f"{n[0]}.{n[1:3]}.{n[3:5]}.{n[5:7]}.{n[7:]}{n[-1]}"
            return act
            break
        
    

print(c_n_account(n))

Solution 2:[2]

You can use creat_number_account[1:3] if you want to print more than 1 character from the input string. Something like this should work for you:

creat_number_account = input("Account Number:")

def c_n_account(creat_number_account):
    if len(creat_number_account) <10:
        print("This account require 10 numbers") 
    elif creat_number_account[10] == "0":
        print("The last number can't be: '0'")
    else:
        print(f'{creat_number_account[0]}.{creat_number_account[1:3]}.{creat_number_account[3:5]}.{creat_number_account[5:7]}.{creat_number_account[7:]}')

You should add a check if the input is bigger than 10 characters.

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1
Solution 2 Sefan