How to Fix Scanner.nextLine() getting skipped after nextInt() in Java

Feb 27, 2026

Problem

When I run a Java program that reads both integer and string input using Scanner.nextLine(), I got this error:

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

Notice how the name input is completely skipped! The program asks for the name but immediately proceeds without letting me type anything.

Environment

Java 21
Windows 11 / macOS / Linux (cross-platform issue)
java.util.Scanner class

What happened?

I was writing a simple Java program that asks for user’s age and name. Here’s my setup:

import java.util.Scanner;

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

        System.out.print("Enter your age: ");
        int age = scanner.nextInt(); // Reads integer from input

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

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

I can explain the key parts:

scanner.nextInt() reads the integer value from input
scanner.nextLine() should read the entire line including the name
Both seem straightforward and should work

But when I run this program and enter 25 for age, the name input gets completely skipped.

How to solve it?

I tried to add a simple scanner.nextLine() after nextInt():

import java.util.Scanner;

public class Solution1 {
    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 newline character

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

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

What changed and why: I added an extra scanner.nextLine() call right after nextInt(). This consumes the newline character left in the buffer.

Now test again:

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

You can see that I succeeded to get both inputs working properly.

The reason

I think the key reason for the error is:

nextInt() only reads the numeric token (25) from the input
It leaves the newline character (\n) in the Scanner’s buffer
The next nextLine() immediately reads this empty line and returns
The program never waits for actual user input for the name

This happens because Scanner’s token-based methods (nextInt(), next(), etc.) work differently from line-based methods (nextLine()). Token-based methods read until they find a delimiter (whitespace, newline, etc.) but don’t consume the delimiter itself.

Summary

In this post, I showed how to resolve Scanner.nextLine() getting skipped when using nextInt() in Java. The key point is understanding Scanner’s buffer behavior with token-based methods versus line-based methods. The simplest fix is to add an extra nextLine() call after nextInt() to consume the newline character, but you can also use Integer.parseInt() with nextLine() for a more consistent approach.

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 API Documentation
👨‍💻 Reddit Java Discussion

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