I can't figure out the "Java InputMismatchException errors" in my code

Question

In my code,I use "\n" as the delimiter, because there may be some 'sapce' in the user's input string. But a exception appeared.I am new to Java and I'm confused.So I'm very grateful to you for helping me out.

My code is:

import java.util.Scanner;

public class ScannerDemo {
    public static void main(String[] args) {

        Scanner scanner=new Scanner(System.in);
        scanner.useDelimiter("\n");

        System.out.print("Please enter your ID:");
        int id=scanner.nextInt();
        System.out.print("Please enter your address:");
        String address=scanner.next();
    }

}

And the output:

Please enter your ID:20151212
Exception in thread "main" java.util.InputMismatchException
    at java.util.Scanner.throwFor(Unknown Source)
    at java.util.Scanner.next(Unknown Source)
    at java.util.Scanner.nextInt(Unknown Source)
    at java.util.Scanner.nextInt(Unknown Source)
    at learning.ScannerDemo.main(ScannerDemo.java:12)

`I use "\n" as the delimiter` This is your problem. – Yassin Hajaj Dec 12 '15 at 14:13 — Yassin Hajaj, Dec 12 '15 at 14:13

score 1 · Accepted Answer · answered Dec 12 '15 at 14:14

1

If you still want to use \n as a delimiter, simply cast the text you're getting after trimming it.

Solution

int id = Integer.valueOf(scanner.next().trim());

Or simply get rid of the delimiter.

answered Dec 12 '15 at 14:14

Yassin Hajaj

21,337
9
51
89

Each time I use both the next() and the nextInt(), I need to get the int in this way?(cast the text,trim it and then convert it) – Kenkon Hu Dec 12 '15 at 14:40
@KenkonHu If you're using `nextInt()` with the delimiter, it will always throw you an exception. If you're not using the delimiter, use these methods normally. :) – Yassin Hajaj Dec 12 '15 at 14:45

score 0 · Answer 2 · answered Dec 12 '15 at 14:17

Setting the delimiter causes the issue - nextInt looks for the "\n" as part of the expected format, but it has already been consumed by the scanner.

If you want a more end-user fiendly approach, you're probably better off scanning for the next string, trim it and convert it to Integer yourself.

score 0 · Answer 3 · answered Dec 12 '15 at 14:25

0

Use scanner.nextLine() then trim() and convert to int.

answered Dec 12 '15 at 14:25

Afshin Ghazi

2,784
4
23
37

score 0 · Answer 4 · answered Dec 12 '15 at 14:56

Since line terminator vary with OS.

Windows system use "\r\n"
Linux system use "\n"

So if you are running your code on windows, you should use "\r\n" as delimiter. But there is a better option provided by java.

System.getProperty("line.separator");

so you should use

scanner.useDelimiter(System.getProperty("line.separator"));

I can't figure out the "Java InputMismatchException errors" in my code

4 Answers4

Solution