Scanner is skipping nextLine() after using next() or nextFoo()? – Dev

The best answers to the question “Scanner is skipping nextLine() after using next() or nextFoo()?” in the category Dev.

QUESTION:

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.

ANSWER:

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 “\n” 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 \n character)
System.out.print("Text1: ");
String text1 = input.nextLine();
System.out.print("Text2: ");
String text2 = input.nextLine();

ANSWER:

That’s because the Scanner.nextInt 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.

You will encounter the similar behaviour when you use Scanner.nextLine after Scanner.next() or any Scanner.nextFoo method (except nextLine itself).

Workaround:

  • Either put a Scanner.nextLine call after each Scanner.nextInt or Scanner.nextFoo to consume rest of that line including newline

    int option = input.nextInt();
    input.nextLine();  // Consume newline left-over
    String str1 = input.nextLine();
    
  • Or, even better, read the input through Scanner.nextLine and convert your input to the proper format you need. For example, you may convert to an integer using Integer.parseInt(String) method.

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

ANSWER:

There seem to be many questions about this issue with java.util.Scanner. I think a more readable/idiomatic solution would be to call scanner.skip("[\r\n]+") to drop any newline characters after calling nextInt().

EDIT: as @PatrickParker noted below, this will cause an infinite loop if user inputs any whitespace after the number. See their answer for a better pattern to use with skip: https://stackoverflow.com/a/42471816/143585

ANSWER:

It’s because when you enter a number then press Enter, input.nextInt() consumes only the number, not the “end of line”. When input.nextLine() executes, it consumes the “end of line” still in the buffer from the first input.

Instead, use input.nextLine() immediately after input.nextInt()