Sentry Answers>Java>

How To Get User Input in Java

How To Get User Input in Java

Venter C.

—

The Problem

I don’t know how to get user input from the console in Java.

The Solution

The most common way to get user input in Java is to use the Scanner class from the java.util package.

For example, the following code uses Scanner with methods such as nextLine() to retrieve different data inputs:

Click to Copy
import java.util.Scanner;

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

        System.out.print("Enter your name: ");
        String name = scanner.nextLine(); // Read a full line of text
        System.out.println("Hello, " + name + "!");

        System.out.print("Enter your age: ");
        int age = scanner.nextInt(); // Read an integer input
        System.out.println("You are " + age + " years old.");

        System.out.print("Enter a decimal number: ");
        double number = scanner.nextDouble(); // Read a double input
        System.out.println("You entered: " + number);

        scanner.close(); // Close the scanner to free resources
    }
}

This demonstrates the following commonly used methods for reading user input:

  • The nextLine() method reads a full line of text.
  • The nextInt() method reads an integer.
  • The nextDouble() method reads a double.

While each method gets a different data type, the Java Scanner class does not offer a native means of reading character input.

If you need to read a single character, you can use nextLine() to fetch a line of text, then extract the first character of that line:

Click to Copy
System.out.print("Enter a character: ");
char ch = scanner.nextLine().charAt(0); // Read first character from input
System.out.println("You entered: " + ch);

You can use try-catch blocks to prevent the program from crashing when invalid input is entered:

Click to Copy
import java.util.InputMismatchException;
import java.util.Scanner;

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

        try {
            System.out.print("Enter an integer: ");
            int number = scanner.nextInt();
            System.out.println("You entered: " + number);
        } catch (InputMismatchException e) {
            System.out.println("Invalid input! Please enter a valid integer.");
        }

        scanner.close();
    }
}

Further Reading

  • Sentry BlogException Handling in Java (with Real Examples)
  • Syntax.fmListen to the Syntax Podcast
  • Syntax.fm logo
    Listen to the Syntax Podcast

    Tasty treats for web developers brought to you by Sentry. Get tips and tricks from Wes Bos and Scott Tolinski.

    SEE EPISODES

Considered “not bad” by 4 million developers and more than 100,000 organizations worldwide, Sentry provides code-level observability to many of the world’s best-known companies like Disney, Peloton, Cloudflare, Eventbrite, Slack, Supercell, and Rockstar Games. Each month we process billions of exceptions from the most popular products on the internet.

learn more about sentryjoin our discord
© 2025 • Sentry is a registered Trademark of Functional Software, Inc.