Getting User input with Scanner

This is because you are using Scanner#next method. And if you look at the documentation of that method, it returns the next token read.

So, when you read user input using next method, it does not read the newline at the end. Which is then read by the nextLine() inside the while loop. And thus, your firstName contains a newline without you knowing.

So, you should use nextLine() in your while rather than next().

Similar is the case with nextInt method. It also does not read the newline. So, you can read using readLine and convert it to int using Integer.parseInt. It can throw NumberFormatException if input value cannot be converted to int. So you need to handle it accordingly.

You can try the below code: –

Scanner sc = new Scanner(System.in);
System.out.println("Continue?[Y/N]");
while (sc.hasNext() && (sc.nextLine().equalsIgnoreCase("y"))) {//change here
    System.out.println("Enter first name");
    String name = sc.nextLine();
    System.out.println("Enter surname");
    String surname = sc.nextLine();
    System.out.println("Enter number");
    int number = 0;
    try {
        number = Integer.parseInt(sc.nextLine());
    } catch (IllegalArgumentException e) {
        e.printStackTrace();
    }
    System.out.println("Continue?[Y/N]");
}

But, note one thing, if you enter a value that cannot be passed to Integer.parseInt you will get an exception, and that input will be skipped. For that case, you need to handle it by using while loop.

Or, if you don’t want to do that exception handling: –

You can add an empty sc.nextLine() after sc.nextInt(), that will consume the newline left over, like this: –

    // Left over part of your while loop

    String surname = sc.nextLine();
    System.out.println("Enter number");
    int number = sc.nextInt();
    sc.nextLine(); // To consume the left over newline;
    System.out.println("Continue?[Y/N]");