How Can I Efficiently Get a File's Extension in Java?

Obtaining File Extensions in Java

If you're looking to retrieve the file extension of a path in Java, you may wonder if there's a straightforward built-in method to do so. While Java doesn't offer a native solution, leveraging Apache Commons IO's FilenameUtils.getExtension method provides a simple and effective alternative.

To utilize FilenameUtils.getExtension, specify the file's full path or its name only. For instance:

import org.apache.commons.io.FilenameUtils;

String ext1 = FilenameUtils.getExtension("/path/to/file/foo.txt"); // "txt"
String ext2 = FilenameUtils.getExtension("bar.exe"); // "exe"

In order to use FilenameUtils.getExtension, you'll need to include the Apache Commons IO dependency in your project:

Maven:

<dependency>
  <groupId>commons-io</groupId>
  <artifactId>commons-io</artifactId>
  <version>2.6</version>
</dependency>

Gradle Groovy DSL:

implementation 'commons-io:commons-io:2.6'

Gradle Kotlin DSL:

implementation("commons-io:commons-io:2.6")

Alternatively, you can refer to the Maven Central repository (https://search.maven.org/artifact/commons-io/commons-io/2.6/jar) for other dependency management options.

The above is the detailed content of How Can I Efficiently Get a File's Extension in Java?. For more information, please follow other related articles on the PHP Chinese website!