Why is the Scanner skipping nextLine after using next or nextFoo

0 votes

We are using the Scanner methods nextInt() and nextLine() for reading input.

It looks like this:

System.out.println("Enter numerical value");    
int option;
option = input.nextInt(); // Read numerical value from input
System.out.println("Enter 1st string"); 
String string1 = input.nextLine(); // Read 1st string (this is skipped)
System.out.println("Enter 2nd string");
String string2 = input.nextLine(); // Read 2nd string (this appears right after reading numerical value)

The problem is that after entering the numerical value, the first input.nextLine() is skipped and the second input.nextLine() is executed so that my output looks like this:

Enter numerical value
3// This is my input
Enter 1st string    // The program is supposed to stop here and wait for my input, but is skipped
Enter 2nd string    // ...and this line is executed and waits for my input

We tested our application and it looks like the problem lies in using input.nextInt(). If we delete it, then both 

string1 = input.nextLine() and string2 = input.nextLine() are executed as I want them to be.

Aug 28, 2018 in Java by Parth
• 4,640 points
1,618 views

1 answer to this question.

0 votes
public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int i = sc.nextInt();
        sc.nextLine();
        double d = sc.nextDouble();
        sc.nextLine();
        String s = sc.nextLine();

        System.out.println("String: " + s);
        System.out.println("Double: " + d);
        System.out.println("Int: " + i);
    }
answered Aug 28, 2018 by samarth295
• 2,220 points

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