What's the Best Way to Read User Input in Java: Scanner Methods Compared?

Feb 27, 2026

Problem

When I wrote a simple Java program to collect both integer and string input, I got this strange behavior:

import java.util.Scanner;

public class InputExample {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        System.out.print("Enter your age: ");
        int age = scanner.nextInt(); // reads 25, leaves \n in buffer
        System.out.print("Enter your name: ");
        String name = scanner.nextLine(); // immediately reads the leftover \n

        System.out.println("Age: " + age);
        System.out.println("Name: '" + name + "'"); // prints empty string!

        scanner.close();
    }
}

The output was:

Age: 25
Name: ''

The name input was completely skipped!

Environment

Java 21
macOS 14.5
Standard Scanner class from java.util

What happened?

I was trying to create a simple program that asks for a user’s age (integer) and name (string). But when I mixed nextInt() and nextLine(), the second input got skipped.

Here’s what I expected to happen:

Enter your age: 25
Enter your name: John
Age: 25
Name: John

But I got an empty string for the name instead.

How to solve it?

I tried several approaches to fix this issue.

First attempt: Buffer management

I realized that nextInt() reads the integer but leaves the newline character (\n) in the buffer. The next nextLine() consumes that leftover newline immediately.

import java.util.Scanner;

public class InputExample {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        System.out.print("Enter your age: ");
        int age = scanner.nextInt();
        scanner.nextLine(); // consume the leftover newline

        System.out.print("Enter your name: ");
        String name = scanner.nextLine();

        System.out.println("Age: " + age);
        System.out.println("Name: '" + name + "'");

        scanner.close();
    }
}

This worked! The output was:

Age: 25
Name: 'John'

Second attempt: Use nextLine() for everything

Then I thought, what if I use nextLine() for all input and parse the integer manually?

import java.util.Scanner;

public class InputExample {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        System.out.print("Enter your age: ");
        int age = Integer.parseInt(scanner.nextLine());

        System.out.print("Enter your name: ");
        String name = scanner.nextLine();

        System.out.println("Age: " + age);
        System.out.println("Name: '" + name + "'");

        scanner.close();
    }
}

This approach also worked and gave me more control over input parsing.

Third attempt: Add validation

But what if the user enters invalid input? I added validation:

import java.util.Scanner;

public class InputExample {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        while (true) {
            System.out.print("Enter your age: ");
            String ageInput = scanner.nextLine();
            try {
                int age = Integer.parseInt(ageInput);
                break;
            } catch (NumberFormatException e) {
                System.out.println("Please enter a valid number");
            }
        }

        System.out.print("Enter your name: ");
        String name = scanner.nextLine();

        System.out.println("Age: " + age);
        System.out.println("Name: '" + name + "'");

        scanner.close();
    }
}

Now the program handles invalid input gracefully.

The reason

I think the key reason for the input skipping problem is that Scanner uses an internal buffer to read input. When you call:

nextInt() - reads characters until it finds an integer, but leaves the delimiter (newline) in the buffer
nextLine() - reads everything until the next newline character, including any leftover newlines

So when you call nextInt() followed by nextLine(), the nextLine() immediately consumes the newline that was left after nextInt().

Summary

In this post, I showed how to resolve the Scanner input skipping problem in Java. The key point is understanding that Scanner methods have different behaviors regarding buffer handling.

Here’s what I learned:

nextInt() leaves the newline character in the buffer
nextLine() consumes the entire line including any leftover newlines
You can either manage the buffer manually or use nextLine() + parsing for better control
Always add input validation to handle invalid user input

The best approach depends on your needs:

Use nextInt() alone when only integers are needed
Use nextLine() + parsing for mixed input scenarios
Always flush the buffer when switching between input types
Add validation to make your code more robust

Final Words + More Resources

My intention with this article was to help others share my knowledge and experience. If you want to contact me, you can contact by email: Email me

Here are also the most important links from this article along with some further resources that will help you in this scope:

👨‍💻 Java Scanner documentation

Oh, and if you found these resources useful, don’t forget to support me by starring the repo on GitHub!