How do I convert scanner input into an integer?

Question

In Java, how, if possible, can I convert numerical scanner input (such as 2 or 87) into an integer variable? What I'm using now yields the error message:

 Exception in thread "main" java.lang.NumberFormatException: null at java.lang.Integer.parseInt(Unknown Source) at java.lang.Integer.valueOf(Unknown Source) at diDecryption.Didecryption.main(Didecryption.java:226)

And this is the code I'm using to do it (pieced together, it's part of a much larger program):

 System.out.println("Enter first number"); Scanner sc=new Scanner(System.in); String name=sc.next(); int result = Integer.valueOf(name); if (result / 2 == 1){ System.out.println("a");

The purpose of the program is to decode an encrypted message. The input is numerical, and if I remove the string to int converter, the division does not work. How do I fix this?

Your code works fine for me if I give it a number. What, exactly, are you providing as input on system in? (The error message you've posted indicates you're trying to parse null) — azurefrog
– azurefrog, Commented Feb 22, 2017 at 15:14
What is your input ? The code looks fine to me. ideone.com/omNdEF — dumbPotato21
– dumbPotato21, Commented Feb 22, 2017 at 15:18
I doubt that your exception message matches your code. Scanner methods don't return null. They will throw an exception if there's nothing to read. Please provide a minimal reproducible example. — Tom
– Tom, Commented Feb 22, 2017 at 15:33

Anik Dey · Accepted Answer · 2017-02-22 18:01:07Z

Try this code

package exmaple; import java.util.Scanner; public class Parser { public static void main(String[] args) { Scanner in = new Scanner(System.in); String name = in.next(); try{ int result = Integer.parseInt(name); if(result / 2 == 1) { System.out.println("a"); } } catch(Exception exception) { } in.close(); } }

FSm · Accepted Answer · 2017-02-22 15:23:25Z

Try

System.out.println("Enter first number"); Scanner sc=new Scanner(System.in); int name=sc.nextInt(); if ((name / 2) == 1) System.out.println("a");

RUN

run: Enter first number 2 a

Alex · Accepted Answer · 2017-02-22 15:11:36Z

System.out.println("Enter first number"); Scanner sc=new Scanner(System.in); String name=sc.next(); int result = Integer.parseInt(name); if (result / 2 == 1){ System.out.println("a");

parseint changes it to a primitive int rather than an Integer object

That won't make any difference, Java's quite happy to unbox the Integer returned by valueOf().

Sergei Stepanenko · Accepted Answer · 2017-02-22 15:21:54Z

In your stacktrace you have null as parameter in Integer.valueOf(name). Seems your console produce some invalid input sequence. Try to check it with sc.hasNext() condition:

 System.out.println("Enter first number"); Scanner sc = new Scanner(System.in); if (sc.hasNext()) { String name = sc.next(); int result = Integer.parseInt(name); if (result / 2 == 1) { System.out.println("a"); } }

David Drezel · Accepted Answer · 2022-10-06 05:43:16Z

try

System.out.println("Enter first number"); Scanner sc=new Scanner(System.in); int result = sc.nextInt; if (result / 2 == 1){ System.out.println("a");

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