error while taking String input in java

Question

package test;
import java.util.*;
public class NewClass {
    public static void main (String[] args)
    {   
        String s;
        Scanner sc = new Scanner (System.in);
        s=sc.nextLine();
        System.out.println(s);
    }

}

I don't know why this error is coming.

Exception in thread "main" java.util.InputMismatchException at java.util.Scanner.throwFor(Scanner.java:864) at java.util.Scanner.next(Scanner.java:1485) at java.util.Scanner.nextInt(Scanner.java:2117) at java.util.Scanner.nextInt(Scanner.java:2076) at test.Test.main(Test.java:12) C:\Users\MOHIT KUMAR SINGH\AppData\Local\NetBeans\Cache\8.2\executor-
snippets\run.xml:53: Java returned: 1 BUILD FAILED (total time: 2 seconds)

This code does not seem to bear much relation to the error in your stack trace. — khelwood, Jun 01 '18 at 11:56
No offense meant, but you could have made some research on your own before asking the whole community for that kind of simple problem. — sjahan, Jun 01 '18 at 11:57
First, your stacktrace does not match your code: in your code, you use `nextLine` though there is only `nextInt` in the stacktrace. It's not the same class name too... Then, you take a look at the doc: https://docs.oracle.com/javase/7/docs/api/java/util/Scanner.html#nextInt() : InputMismatchException - if the next token does not match the Integer regular expression, or is out of range. This explains the exception. — sjahan, Jun 01 '18 at 12:02
Possible duplicate of [Why am I getting InputMismatchException?](https://stackoverflow.com/questions/14027537/why-am-i-getting-inputmismatchexception) — sjahan, Jun 01 '18 at 12:08
@MohitKumarsingh I think you do not get it. As I already told you: the code and the stacktrace does not match. The error is caused by a call to `nextInt` which does not exist in your code, which means you don't actually show your code... Or you don't know what you actually run. **Once again, `nextLine` does not throw InputMismatchException** — sjahan, Jun 01 '18 at 12:19
Ok, I can confirm what I just said: you don't know what you actually run. Here, this is the `Test` class you run, not the `NewClass`. — sjahan, Jun 01 '18 at 12:24

score 0 · Accepted Answer · answered Jun 01 '18 at 12:06

0

java.util.InputMismatchException-In order to deal with this exception you must verify that the input data of your application meet its specification. When this error is thrown, the format of the input data is incorrect and thus, you must fix it, in order for your application to proceed its execution.

answered Jun 01 '18 at 12:06

Sweta Patra

341
1
4

but I am giving string as an input. – Mohit Kumar singh Jun 01 '18 at 12:08
@MohitKumarsingh if you are giving string as input then You should not get this error. Even if you give any symbol or integer as input then It should print without any error due to string input. – Nitin Bisht Jun 01 '18 at 12:09
but I am getting this error . – Mohit Kumar singh Jun 01 '18 at 12:11
@MohitKumarsingh Please put your complete code here alongwith complete error. – Nitin Bisht Jun 01 '18 at 12:11
package test; import java.util.*; public class NewClass { public static void main (String[] args) { String s; Scanner sc = new Scanner (System.in); s=sc.nextLine(); System.out.println(s); } } – Mohit Kumar singh Jun 01 '18 at 12:12
@MohitKumarsingh not here in comment. Edit your question and add updated question – Nitin Bisht Jun 01 '18 at 12:14
this is complete code. – Mohit Kumar singh Jun 01 '18 at 12:14
@MohitKumarsingh how you are running this code. – Nitin Bisht Jun 01 '18 at 12:16
I using netBeans and running by pressing that green triangle button – Mohit Kumar singh Jun 01 '18 at 12:19
@MohitKumarsingh I have copied and tested your code exactly. It is working fine in eclipse. Please check you must be having some other issue. – Nitin Bisht Jun 01 '18 at 12:20
@NitinBisht Ok thank you. – Mohit Kumar singh Jun 01 '18 at 12:24

score 0 · Answer 2 · answered Jun 01 '18 at 12:07

0

I think it's your main class but there no argument/parameter in your main method. You should do like this

public static void main(String[] arr) {
    Scanner sc = new Scanner(System.in);
    String s = sc.nextLine();
    System.out.println(s);
}

answered Jun 01 '18 at 12:07

Dhirendra Sharma

132
1
6

I did corrected this thing but still nothing. – Mohit Kumar singh Jun 01 '18 at 12:09
You debug your code..? – Dhirendra Sharma Jun 01 '18 at 12:10

error while taking String input in java

2 Answers2