Using scanner.nextLine()

Taylor P. picture Taylor P. · Feb 17, 2011 · Viewed 683.5k times · Source

I have been having trouble while attempting to use the nextLine() method from java.util.Scanner.

Here is what I tried:

import java.util.Scanner;

class TestRevised {
    public void menu() {
        Scanner scanner = new Scanner(System.in);

        System.out.print("Enter a sentence:\t");
        String sentence = scanner.nextLine();

        System.out.print("Enter an index:\t");
        int index = scanner.nextInt();

        System.out.println("\nYour sentence:\t" + sentence);
        System.out.println("Your index:\t" + index);
    }
}

Example #1: This example works as intended. The line String sentence = scanner.nextLine(); waits for input to be entered before continuing on to System.out.print("Enter an index:\t");.

This produces the output:

Enter a sentence:   Hello.
Enter an index: 0

Your sentence:  Hello.
Your index: 0

// Example #2
import java.util.Scanner;

class Test {
    public void menu() {
        Scanner scanner = new Scanner(System.in);

        while (true) {
            System.out.println("\nMenu Options\n");
            System.out.println("(1) - do this");
            System.out.println("(2) - quit");

            System.out.print("Please enter your selection:\t");
            int selection = scanner.nextInt();

            if (selection == 1) {
                System.out.print("Enter a sentence:\t");
                String sentence = scanner.nextLine();

                System.out.print("Enter an index:\t");
                int index = scanner.nextInt();

                System.out.println("\nYour sentence:\t" + sentence);
                System.out.println("Your index:\t" + index);
            }
            else if (selection == 2) {
                break;
            }
        }
    }
}

Example #2: This example does not work as intended. This example uses a while loop and and if - else structure to allow the user to choose what to do. Once the program gets to String sentence = scanner.nextLine();, it does not wait for input but instead executes the line System.out.print("Enter an index:\t");.

This produces the output:

Menu Options

(1) - do this
(2) - quit

Please enter your selection:    1
Enter a sentence:   Enter an index: 

Which makes it impossible to enter a sentence.


Why does example #2 not work as intended? The only difference between Ex. 1 and 2 is that Ex. 2 has a while loop and an if-else structure. I don't understand why this affects the behavior of scanner.nextInt().

Answer

Peter Lawrey picture Peter Lawrey · Feb 17, 2011

I think your problem is that

int selection = scanner.nextInt();

reads just the number, not the end of line or anything after the number. When you declare

String sentence = scanner.nextLine();

This reads the remainder of the line with the number on it (with nothing after the number I suspect)

Try placing a scanner.nextLine(); after each nextInt() if you intend to ignore the rest of the line.