How to run a bat file in Startup directory with Java?

Question

I tried both of the following:

Runtime.getRuntime().exec("cmd.exe /c C:/Users/USER/AppData/Roaming/Microsoft/Windows/Start Menu/Programs/Startup/MyApp.bat");

Runtime.getRuntime().exec("cmd.exe /c \"C:/Users/USER/AppData/Roaming/Microsoft/Windows/Start Menu/Programs/Startup/MyApp.bat\"");

Neither of them worked, the first one didn't have any error message, the second one had the following error message:

java.io.IOException: Cannot run program "cmd.exe /c C:/Users/USER/AppData/Roaming/Microsoft/Windows/Start Menu/Programs/Startup/MyApp.bat": CreateProcess error=2, The system cannot find the file specified at java.lang.ProcessBuilder.start(ProcessBuilder.java:1048)

MyApp.bat is in the Startup directory and I can run it by hand.

What's the correct way to run it from my Java app?

Answer 1

For opening any file on the computer - the Desktop class is a great fit. Here's how it can be implemented:

import java.awt.*;
import java.io.*;
public class OpenBat {
    public static void main(String[] args) throws IOException {
        Desktop desktop = Desktop.getDesktop();
        File bat = new File("C:/Users/USER/" +
        "AppData/Roaming/Microsoft/Windows/Start " + 
        "Menu/Programs/Startup/MyApp.bat");

        desktop.open(bat);

    }
}

Answer 2

好的，我知道了，它是：

Runtime.getRuntime().exec("cmd /C start \"\" \"C:/Users/USER/AppData/Roaming/Microsoft/Windows/Start Menu/Programs/Startup/MyApp.bat\"");