How do I get the file extension of a file in Java?

Do you really need a "parser" for this?

String extension = "";

int i = fileName.lastIndexOf('.');
if (i > 0) {
    extension = fileName.substring(i+1);
}

Assuming that you're dealing with simple Windows-like file names, not something like archive.tar.gz.

Btw, for the case that a directory may have a '.', but the filename itself doesn't (like /path/to.a/file), you can do

String extension = "";

int i = fileName.lastIndexOf('.');
int p = Math.max(fileName.lastIndexOf('/'), fileName.lastIndexOf('\\'));

if (i > p) {
    extension = fileName.substring(i+1);
}

In this case, use FilenameUtils.getExtension from Apache Commons IO

Here is an example of how to use it (you may specify either full path or just file name):

import org.apache.commons.io.FilenameUtils;

// ...

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

Maven dependency:

<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")

Others https://search.maven.org/artifact/commons-io/commons-io/2.6/jar

private String getFileExtension(File file) {
    String name = file.getName();
    int lastIndexOf = name.lastIndexOf(".");
    if (lastIndexOf == -1) {
        return ""; // empty extension
    }
    return name.substring(lastIndexOf);
}