Reading a string and adding it to an ArrayList

Question

I wrote a program that has an arraylist of users so it should either create a new user and store the user's name in an arraylist or choose an existing user from an arraylist. The problem is, when I run the code I can choose an option but when I want to create a new user I cannot type anything. Why does that happen and how can I fix this?

import java.util.ArrayList;
import java.util.Iterator;
import java.util.Scanner;

public class Main {

private ArrayList<String> users;

public static void main(String[] args) {
    Main m = new Main();
    m.menu();
}

public void createNewUser(String name) {
    users.add(name);
}

public boolean search(String username) {
    if (users.contains(username))
        return true;
    else
        return false;
}

public void displayUsers() {
    Iterator<String> itr = users.iterator();
    while (itr.hasNext()) {
        String name = itr.next();
        System.out.println(name);
    }
}

public void menu() {
    users = new ArrayList<String>();
    int choice = 0;
    String name;
    Scanner input = new Scanner(System.in);

    System.out.println("Choose an option:");
    System.out.println("1- Create new user");
    System.out.println("2- Select current user");

    choice = input.nextInt();

    if (choice == 1) {
        System.out.println("Enter your name: ");
        name = input.nextLine();
        createNewUser(name);
        displayUsers();
    } else {
        do {
            displayUsers();
            System.out.println("Enter your username: ");
            name = input.nextLine();
            if (search(name)) {
                System.out.println("User found");
                break;
            }
            else
                System.out.println("This username does not exist.");
        } while (!search(name));

    }

}
}

Well known behavior you must consume the newline [Check this answer](http://stackoverflow.com/questions/13102045/skipping-nextline-after-use-nextint) — gtgaxiola, Oct 01 '14 at 18:14
Why use `if`-`else` in the `search()` method; you can just have `return users.contains(username)`, which eliminates the need for a separate method from `contains()` in the first place... — bcsb1001, Oct 01 '14 at 18:17

score 0 · Answer 1 · answered Oct 01 '14 at 18:16

0

System.out.println("Choose an option:");
    System.out.println("1- Create new user");
    System.out.println("2- Select current user");

    choice = input.nextInt();
    input.nextLine();

Try adding input.nextLine(); after your choice.

answered Oct 01 '14 at 18:16

brso05

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While this may solve the problem it is good to also provide explanation why problem occurred and why your solution corrects it so OP and other readers could learn from it. – Pshemo Oct 01 '14 at 18:25
I tried doing that and it didn't work – Oct 01 '14 at 18:28
It should work it works for me what problem did you have? Error message? – brso05 Oct 01 '14 at 18:29
No error message, I cannot type anything as before. – Oct 01 '14 at 18:31
Are you sure you saved and recompiled? – brso05 Oct 01 '14 at 18:31
I copied and pasted your code and added that line and it works fine for me. – brso05 Oct 01 '14 at 18:31
Yes I'm sure, is it an issue with my computer? – Oct 01 '14 at 18:34
No java runs on a virtual machine it shouldn't have anything to do with your computer. – brso05 Oct 01 '14 at 18:34
Are you using eclipse, netbeans, command prompt? – brso05 Oct 01 '14 at 18:34
I tried in eclipse and via command prompt both worked for me. – brso05 Oct 01 '14 at 18:40
I noticed your not doing anything after entering name other than printing out then program exits. – brso05 Oct 01 '14 at 18:40
Will it not even let you type at all? – brso05 Oct 01 '14 at 18:41

Reading a string and adding it to an ArrayList

1 Answers1