Trying to sort user input into alphabetical order with Java, why won't this code work?

Question

For some reason when I try to ask a user for names so I can add them to a list and sort them alphabetically, this code will not print anything out. It will not even get past the while loop, does anyone have any idea what the problem is? Also another question; how can you execute some code if the user presses the enter button when asked for input value, would it just be null? Thanks!

import java.util.Scanner;
import java.util.*;

public class project16u
{
    public static void main(String[] args) 
    {
        int n;
        String input = "nothing";
        String temp;
        ArrayList<String> names = new ArrayList<String>();
        Scanner s1 = new Scanner(System.in);
        System.out.println("Enter all the names:");
        while(!input.equals("done")){
            input = s1.nextLine();
            names.add(input);
        }
        for (int i = 0; i < names.size()-1; i++) 
        {

                if (names.get(i).compareTo(names.get(i+1))>0) 
                {
                    temp = names.get(i);
                    names.add(i, names.get(i+1));
                    names.add(i+1, temp);
                    i=0;
                }
        }
        System.out.print("Names in Sorted Order:");
        for (int i = 0; i < names.size() - 1; i++) 
        {
            System.out.print(names.get(i).toString() + ",");
        }
        System.out.print(names.get(names.size()-1));
    }
}

You'll need to debug your code. It surely stays within the for loop because you are resetting i=0 at some point within it, which is generally a bad practice: All loops should iterate once and just once on every value of the control variable. They might exit promptly, but not lately. — Little Santi
– Little Santi, Commented Mar 26, 2017 at 17:22

Joe C · Accepted Answer · 2017-03-26 17:16:52Z

1

add inserts the name at the requested index. Thus, in your case, you will have two copies of the same name in the list, rather than the one you intended.

You probably want to use set instead.

answered Mar 26, 2017 at 17:16

Joe C

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OTM · Accepted Answer · 2017-03-26 17:38:19Z

0

You may need to change the loop condition to

s1.hasNextLine() && !input.equals("done")

answered Mar 26, 2017 at 17:38

OTM

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