Scanner is skipping nextLine after using next or nextFoo

675    Asked by RickiGriffen in Java , Asked on Jul 26, 2021

 I am 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

I tested my application and it looks like the problem lies in using input.nextInt(). If I delete it, then both string1 = input.nextLine() and string2 = input.nextLine() are executed as I want them to be.

Answered by Gillian Hamer

The problem is with the input.nextInt() method - it only reads the int value. So when you continue reading with input.nextLine() you receive the "
" Enter key. So to skip this you have to add the input.nextLine(). Hope this should be clear now.

Try it like that:

System.out.print("Insert a number: ");
int number = input.nextInt();
input.nextLine(); // This line you have to add (It consumes the
character)
System.out.print("Text1: ");
String text1 = input.nextLine();
System.out.print("Text2: ");
String text2 = input.nextLine();

Note: java.nextIine method does not read the newline character in your input created by hitting "Enter," and so the call to Scanner. nextLine returns after reading that newline.



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