Run .exe file in Java from file location

Question

I have to open a .exe file from my Java program. So I tried following code First.

Process process = runtime.exec("c:\\program files\\test\\test.exe");

But I was getting some error. Then I found out that the exe has to be launched from that location that is c://program files/test/ only then it will open with out errors. So I decided to write a .bat file and execute so that it will cd to that location and execute the .exe file.

Following is my code:

BufferedWriter fileOut;

String itsFileLocation = "c:\\program files\\test\\"
    System.out.println(itsFileLocation);
    try {
     fileOut = new BufferedWriter(new FileWriter("C:\\test.bat"));
     fileOut.write("cd\\"+"\n");
     fileOut.write("cd "+ itsFileLocation +"\n");
     fileOut.write("test.exe"+"\n");
     fileOut.write("exit"+"\n");
     
     fileOut.close(); // Close the output stream after all output is done.
    } catch (IOException e1) {
     e1.printStackTrace();
    } // Create the Buffered Writer object to write to a file called filename.txt
    Runtime runtime = Runtime.getRuntime();
    try {
     Process process =runtime.exec("cmd /c start C:\\test.bat");
    } catch (IOException e) {
     e.printStackTrace();
    }

The above code works perfectly. However, the command prompt is also opened at the back of my .exe (Application). It closes only after the .exe file exits..

I need to clse my command prompt when my application stats.

My .bat file will be like following after it is written by the program.

cd\
cd C:\Program Files\test\
test.exe
exit

Answer 1

You don't need a console. You can execute a process using a working directory:

exec(String command, String[] envp, File dir)

Executes the specified string command in a separate process with the specified environment and working directory.

• command is the location of the .exe
• envp can be null
• dir, is the directory of your .exe

With respect to your code it should be...

Runtime.getRuntime().exec("c:\\program files\\test\\test.exe", null, new File("c:\\program files\\test\\"));

Answer 2

You can use Runtime.exec(java.lang.String, java.lang.String[], java.io.File) where you can set the working directory.

Or else you can use ProcessBuilder as follows:

ProcessBuilder pb = new ProcessBuilder("myCommand", "myArg1", "myArg2");
pb.directory(new File("myDir"));
Process p = pb.start();

Answer 3

Another way of running a file is the following:

import java.awt.Desktop;
import java.io.File;

public static void open(String targetFilePath) throws IOException
{
    Desktop desktop = Desktop.getDesktop();

    desktop.open(new File(targetFilePath));
}

Answer 4

The the Standard Code for Running bat or any other through command line using java is :

runtimeProcess = Runtime.getRuntime().exec("cmd /c start cmd.exe /C\""+backup_path+"\"");
int processComplete = runtimeProcess.waitFor();

and you can go on continue for multiple files using & supperator like : &&

Answer 5

This would also work.

 Process process = new ProcessBuilder("C:\\Users\\test\\Downloads\\Termius.exe").start();

It would start the .exe in that file location.

Answer 6

Best way to run exe file

make java.awt.Desktop object and equal Desktop.getDesktop();

Desktop desktop = Desktop.getDesktop(); desktop.open("file path");

run exe file:

desktop.open("C:\\Windows\\System32\\cmd.exe");

or

desktop.open("C:/Windows/System32/cmd.exe");

run url :

desktop.browse(new URI("http://www.google.com"));