在java中打开Windows资源管理器

Question

I have been looking for an answer to this on Stack Overflow, but I couldn't find an answer that worked for me.

Using Java, how do I create a button that will launch an Explorer Window to a specified directory? If this is possible, how do I make it work for OSX and Linux?

Answer 1

I am not sure how it works in other OS but in Windows you can use something like this

Desktop.getDesktop().open(new File("c:\\"));

Edit

Found another way (check link to FileExplorer class from that answer). Also you can use System.getProperty("os.name") to determine operation system.

Answer 2

javax.swing.JButton myButton = new javax.swing.JButton("BUTTON TEXT");
myButton.addActionListener(new java.awt.event.ActionListener() {

  @Override
  public void actionPerformed(ActionEvent e) {
    java.awt.Desktop.getDesktop().open(new java.io.File("MY PATH NAME HERE"));
  }
});