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);
}