How to Validate UUID Strings in Java

Universally Unique Identifiers (UUIDs) are widely used in applications for uniquely identifying entities such as database records, transactions, and distributed objects. In Java, validating a UUID string is essential to ensure it conforms to the correct format and prevents errors in your application.

This article will guide you through different methods for validating UUIDs in Java using built-in and third-party libraries.

What is a UUID?

A UUID is a 128-bit value typically represented as a 36-character string in the format:

xxxxxxxx-xxxx-Mxxx-Nxxx-xxxxxxxxxxxx

where:

• x represents hexadecimal digits (0-9, a-f).
• M represents the version (1, 2, 3, 4, or 5).
• N represents the variant (8, 9, A, or B).

Method 1: Using Java’s Built-in UUID Class

Java provides a built-in UUID class that can be used to validate UUID strings by attempting to parse them.

import java.util.UUID;

public class UUIDValidator {
    public static boolean isValidUUID(String uuid) {
        try {
            UUID.fromString(uuid);
            return true;
        } catch (IllegalArgumentException e) {
            return false;
        }
    }

    public static void main(String[] args) {
        String testUUID = "550e8400-e29b-41d4-a716-446655440000";
        System.out.println("Is valid UUID? " + isValidUUID(testUUID));
    }
}

Explanation

• UUID.fromString(uuid) attempts to parse the input string.
• If the string is a valid UUID, it returns a UUID object.
• If the string is invalid, an IllegalArgumentException is thrown, and we return false.

Method 2: Using Regular Expressions

A regular expression (regex) can be used to check if a string conforms to the standard UUID format.

import java.util.regex.Pattern;

public class UUIDRegexValidator {
    private static final String UUID_REGEX =
            "^[0-9a-fA-F]{8}-[0-9a-fA-F]{4}-[1-5][0-9a-fA-F]{3}-[89abAB][0-9a-fA-F]{3}-[0-9a-fA-F]{12}$";
    private static final Pattern UUID_PATTERN = Pattern.compile(UUID_REGEX);

    public static boolean isValidUUID(String uuid) {
        return uuid != null && UUID_PATTERN.matcher(uuid).matches();
    }

    public static void main(String[] args) {
        String testUUID = "550e8400-e29b-41d4-a716-446655440000";
        System.out.println("Is valid UUID? " + isValidUUID(testUUID));
    }
}

Explanation

• The regex pattern ensures the UUID follows the standard format.
• It checks for a valid version (1-5) and variant (8, 9, A, or B).
• Pattern.matcher(uuid).matches() verifies if the input string conforms to the pattern.

Method 3: Using Apache Commons Validator

Apache Commons Validator provides a utility for checking UUID validity.

import org.apache.commons.validator.routines.RegexValidator;

public class ApacheUUIDValidator {
    private static final RegexValidator UUID_VALIDATOR = new RegexValidator(
            "^[0-9a-fA-F]{8}-[0-9a-fA-F]{4}-[1-5][0-9a-fA-F]{3}-[89abAB][0-9a-fA-F]{3}-[0-9a-fA-F]{12}$");

    public static boolean isValidUUID(String uuid) {
        return UUID_VALIDATOR.isValid(uuid);
    }

    public static void main(String[] args) {
        String testUUID = "550e8400-e29b-41d4-a716-446655440000";
        System.out.println("Is valid UUID? " + isValidUUID(testUUID));
    }
}

Explanation

• Uses RegexValidator from Apache Commons Validator to check UUID format.
• Provides a robust and reusable solution.
• Requires adding the Apache Commons Validator dependency:

<dependency>
    <groupId>commons-validator</groupId>
    <artifactId>commons-validator</artifactId>
    <version>1.9</version>
</dependency>

Conclusion

Validating UUIDs in Java can be done using different approaches:

1. Built-in UUID class: Simple and effective but doesn’t strictly enforce version and variant rules.
2. Regular Expressions: More control over validation rules but slightly more complex.
3. Apache Commons Validator: A third-party solution that provides a cleaner way to validate UUIDs.

Choose the method that best suits your application’s requirements. If you’re dealing with standard UUID parsing, the UUID class is usually sufficient, but for stricter validation, regex or Apache Commons Validator can be useful.

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Anastasios Antoniadis

Editor-in-Chief at Beyond Tech Now

Anastasios Antoniadis is the Editor in Chief of Beyond Tech Now, where he leads a diverse team of writers and analysts delivering industry-breaking news, hands-on product reviews, and thoughtful commentary on emerging technologies. With over a decade of experience in digital media, John has a proven track record of shaping editorial vision and driving engagement through high-quality, reader-focused content. He is particularly passionate about the intersection of AI, cybersecurity, and consumer electronics—striving to make complex tech topics accessible to a global audience. When he’s not collaborating with writers or liaising with tech leaders, Anastasios can be found experimenting with the latest gadgets or sharing insights at tech conferences around the world.

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