如何在启动时将我的java应用程序（如.exe）传递给文档的路径？

Question

I use JSmooth to turn my java application (jar) into an executable. It´sa simple text editor, like notepad. I want to use the "Open With" function of windows to open certain files with the exe. For this, I only need the path of each file. How do I do this?

I thought about using a java-property and calling it with System.getProperty("VariableName"), though I don´t know if this is possible. ${EXECUTABLEPATH} just gets me the location of the MyApp.exe.

Answer 1

For short properties may use Preferences.userNodeForPackage(MyExample.class); and methods: get, put, flush.
https://docs.oracle.com/javase/8/docs/technotes/guides/preferences/index.html

and/or to use properties file near your exe file for save paths https://docs.oracle.com/javase/7/docs/api/java/util/Properties.html

For open with Windows may use this schema for example:

if (Desktop.isDesktopSupported()) {
    Desktop desktop = Desktop.getDesktop();
    if (desktop.isSupported(Desktop.Action.OPEN)) {
         desktop.open(file);
    }
}

https://docs.oracle.com/javase/7/docs/api/java/awt/Desktop.html

For set associations see: https://www.thewindowsclub.com/change-file-associations-windows may be usefull Have the ability to set Java application as default file opener?