Home > DeveloperSection > Forums > Java Scanner issue : Why input is skipped when using next(), nextInt() etc. after nextLine()
Mary Johnson
Mary Johnson

Total Post:3

Posted on    October-09-2015 6:13 AM

 Java Java 

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I am working on a small java implementation, but  i am astonished  when i saw a weird behavior of scanner class methods, If i need to take two inputs  : first one integer value and next a String. So i used nextInt() first for integer and then nextLine() for string :
import java.util.Scanner;

public class ScannerIssue {

public static void main(String[] args){

Scanner sc = new Scanner(System.in);
System.out.println("Enter an integer");
int i = sc.nextInt();
System.out.println("Enter a String");
String j = sc.nextLine();

Now the output is 
Enter an integer
Enter a String
i.e it skipped the input from nextLine() method and only prompting input once for integer and print its value 
What is going wrong in this code?

Mayank Tripathi
Mayank Tripathi

Total Post:397

Posted on    October-09-2015 6:50 AM

Hi Mary,

I think its kind of a unnoticeable bug in scanner class, here the scanner's nextInt() method doesn't consumes the last newline character of the input , which is then consumed in the next call to scanner nextLine() method.

We can handle this by  calling a blank nextLine() after nextInt() to consume the left over new line chaacter
int option = input.nextInt();
input.nextLine();  // Consume newline left-over
String str1 = input.nextLine();
We can also do this, by taking both the inputs as String by calling nextLine() method and convert the integer value using Integer.parseInt(string) method. For this we need to use try-catch block coz Integer.parseInt throws NumberFormatException when an invalid argument pass through it. 

int option = 0;
try {
    option = Integer.parseInt(input.nextLine());
} catch (NumberFormatException e) {
String str1 = input.nextLine();

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