Java Menu Loop Problems; Program loops too soon?

Question

I am working on an assignment where I have to write a program to encrypt and decrypt Caeser ciphers. The part I am having trouble with though is not encrypting or decrypting, but another one of the requirements, which was that I have to provide a menu so that the user can choose to encrypt, decrypt, or quit. Furthermore, the program should keep prompting the use until the use selects quit. My code so far is:

import java.util.*;
public class CaeserShiftTester
{
    public static void main (String[] args)
    {
        Scanner in = new Scanner(System.in);
        String choice = "";

        while (!choice.equalsIgnoreCase("C"))
        {

        System.out.println("\nPlease select an option");
        System.out.println("[A] Encrypt Code");
        System.out.println("[B] Decrypt Code");
        System.out.println("[C] Quit");

        choice = in.next();

        if(choice.equalsIgnoreCase("A"))
        {
            System.out.println("Please enter your key:");

            final int KEY = in.nextInt();
            System.out.println(CaeserShiftEncryption.shiftAlphabet(KEY));

            System.out.println("\nPlease enter your message:");
            String message = in.nextLine();

            System.out.println(CaeserShiftEncryption.encryptCode(message,KEY));


        }

        if(choice.equalsIgnoreCase("C"))
        {
            System.out.println();
        }
    }


    }

}

My problem is, after the "New Alphabet" is printed out to the screen, the program loops back to the very beginning, asking the user to choose a, b, or c. The use never gets a chance to enter a message to be encrypted. Unfortunately, I am required to print out the New Alphabet that is generated, and I can't think of what might be wrong here. I hope you guys can help me out.

Also, the shiftAlphabet and encryptCode methods are both fully functional.

Looping back to the beginning should ask the user for choices, right? Otherwise the user has no way to decrypt after encrypting something or vice versa. — Jimmt
– Jimmt, Commented Dec 9, 2012 at 4:22
Put System.out.println(choice); in the first if body and see what you get. — DrinkJavaCodeJava
– DrinkJavaCodeJava, Commented Dec 9, 2012 at 4:27

Blaskovicz · Accepted Answer · 2012-12-09 04:37:25Z

1

Take a look on this website: http://www.java-made-easy.com/java-scanner.html

particularly

don't try to scan text with nextLine(); AFTER using nextInt() with the same scanner! It doesn't work well with Java Scanner, and many Java developers opt to just use another Scanner for integers

answered Dec 9, 2012 at 4:37

Blaskovicz

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1 Comment

user1888863 Over a year ago

Thanks! I didn't even notice that I used the same scanner for both of them. The program is working fine now.

Duane · Accepted Answer · 2012-12-10 12:04:57Z

0

Problems occur with the Scanner class when mixing different type of scans and then using the same Scanner instance with nextLine.

When you get such a problem try to create a new Scanner instance. A simple cure would be to have 1 Scanner for using nextLine and another for everything else.

In the past I've had problems with using the Scanner when it came to next and nextLine, the cure for me was to simply stick to only using nextLine.

answered Dec 10, 2012 at 12:04

Duane

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