How to open file explorer in java

Java Open File Explorer

Sometimes, we come across requirements where we need to open a particular folder or directory in the native file explorer. In this tutorial, we will see how to do that, easy and effective.

Opening file explorer with java in any platform – Windows, Linux and Mac

The easiest way to do this is by using Java Desktop. The class provides an open() function Desktop#open(File) to open explorer when the given file is a directory.

private static void openDirectory() throws IOException {
  File directory = new File("C://Program Files//");
  Desktop.getDesktop().open(directory);
}

Open file explorer and select/highlight a file

Now, if you want to open the file explorer and have a file preselected, then there is a way for that as well! You can make use of Desktop.fileBrowseDirectory(File) to launch the file explorer and have a file highlighted. Let’s see a code example for the same.

However, before using this function, you have to keep the following things in mind.

  • The code will not work on ALL platforms.
  • You need to check whether the feature is supported via Desktop.getDesktop().isSupported(Action.BROWSE_FILE_DIR) function.
  • Minimum Java 9 is required
import java.awt.Desktop;
import java.awt.Desktop.Action;
import java.io.File;

public class JavaFileExplorer {

  public static void main(String[] args) {
    openDirectory();
  }

  private static void openDirectory() {
    //Check if the feature supported on your platform
    if (Desktop.getDesktop().isSupported(Action.BROWSE_FILE_DIR)) {
      File directory = new File("/home/afsal/Downloads/jdk-8u212-linux-x64.tar.gz");
      //Open directory with browse_file_dir option
      Desktop.getDesktop().browseFileDirectory(directory);
    }
  }
}

Open file explorer in Linux

Now, if you would like to add specific code for Linux/Ubuntu to open the file explorer, have a look into the following code snippet. We can make use of the xdg-open command to open any directory. The good thing about using this method is that, it works across multiple Linux distributions. They just have to have the xdg-open support.

private static void openDirectoryInLinux() throws Exception {
  Runtime.getRuntime().exec(
     new String[]{"sh", "-c", "/usr/bin/xdg-open '/home/genuinecoder/Downloads/'"}
   );
}

Open file explorer in Windows

Now, if you would like to add specific code for Windows OS to open the file explorer, have a look into the following code snippet. We will execute a cmd command to open Windows explorer.exe.

private static void openDirectoryInWindows() throws Exception {
  Runtime.getRuntime().exec("explorer C:\\");
}

Open file explorer in MacOS

If you are looking for specific code for macOS, you can make use of the /usr/bin/open function to open the directory on macOS.

private static void openDirectoryInMac() throws Exception {
   Runtime.getRuntime().exec(new String[]{"/usr/bin/open", file.getAbsolutePath()});
}

Conclusion

In this tutorial, we have learned how to open a directory from a java program. We have numerous methods to achieve the same goal here. We can use java.awt.Desktop for platform independent support and if platform-specific implementation is required, that is also available. You might also be interested in the following topics.

  1. Java TrayIcon Tutorial
  2. Directory / Folder Indexing Program Using Java
Muhammed Afsal Villan
Muhammed Afsal Villan is an experienced full-stack developer, specialized in desktop and mobile application development. He also regularly publishes quality tutorials on his YouTube channel named 'Genuine Coder'. He likes to contribute to open-source projects and is always enthusiastic about new technologies.

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