Thread: Opening a file with a java program

This is an OS problem as opposed to a java problem. For each OS you need to associate the file with the application, which can be relatively easy or relatively hard depending upon the OS. Alternatively, a user could define the application to open the file with using a simple right click and defining your application.

I've managed to associate the .jar with a file, but the program does not launch. Here's some code.

package simpleImageViewer;
 
import javax.swing.ImageIcon;
import javax.swing.JFrame;
import javax.swing.JLabel;
import javax.swing.JPanel;
import javax.swing.SwingUtilities;
 
public class Main extends JFrame{
 
	public String source = null;
	public JPanel stage = null;
 
	public Main(String src) {
		source = src;
 
		add(stage = new JPanel());
		stage.add(new JLabel(new ImageIcon(source)));
 
		this.pack();
		this.setDefaultCloseOperation(EXIT_ON_CLOSE);
		this.setLocationRelativeTo(null);
		this.setVisible(true);
	}
 
    public static void main(final String[] args) {
		SwingUtilities.invokeLater(new Runnable() {
			public void run() {
				Main m = new Main(args[0]);
			}
		});
    }
 
}

I think, perhaps, I am making assumptions on what is being passed as the parameter to main(). I've assumed a path to the file is being passed, but this probably isn't the case, is it?

Sorry to bump, but I still haven't figured this out yet.

To reiterate, what I'm trying to do is to open a file by dragging it over MyJavaApp.jar, or in windows (MyFile > open with > MyJavaApp.jar).
What I've assumed was to happen is that the file path is passed to the application as an argument to the main method, but my app doesn't even start up.

Ok, I was looking into batch scripts and I found out I can do this:

"C:\Program Files\Java\jre6\bin\javaw.exe" -Xms64M -Xmx4G -jar "C:\path\to\myApp.jar" %*

I just open my file with the .bat file, which then passes it to the .jar.
This solution is good enough for my current needs