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

Schifty · Accepted Answer · 2012-05-21 14:57:21Z

29

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\\"));

edited May 21, 2012 at 14:57

answered May 21, 2012 at 13:18

Schifty

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4 Comments

Dunes Over a year ago

Java will correctly interpret / as a file separator on windows -- at least on windows 7.

Kuldeep Jain Over a year ago

@schifty: In fact we do not need to give full path of test.exe in the first parameter of exec(String command, String[] envp, File dir). So essentially Runtime.getRuntime().exec("test.exe", null, new File("c:\\program files\\test\\")); is good enough.

Nom1fan Over a year ago

@KuldeepJain That's not entirely true and a bit misleading. If the exe is in PATH then it's true full path is not needed, but otherwise, it is.

Kuldeep Jain Over a year ago

@Nom1fan, Looks like you did not follow my comment well. The path can be specified in the 3rd parameter. I mentioned to not give full path in 1st parameter. Please see my example I gave in comment. You can also read further at: docs.oracle.com/javase/7/docs/api/java/lang/…

Kuldeep Jain · Accepted Answer · 2012-05-21 13:16:28Z

13

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();

answered May 21, 2012 at 13:16

Kuldeep Jain

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6 Comments

Dilip Rajkumar Over a year ago

Thanks Kuldeep I have already tried this.. it is not working I am getting the same error. It is expecting me to be in the same directory of .exe file when I launch the .exe file

Dilip Rajkumar Over a year ago

If possible can you give the Runtime.exec example.. with respect to my code.. thanks in advance..

Dilip Rajkumar Over a year ago

Thanks Kuldeep, I have also tried ProcessBuilder.. I am getting the same problem..

Dilip Rajkumar Over a year ago

The .exe files need some property file from that location to start.. I can do one thing is I can set that path as env variable.. I am not sure will that work..

Dunes Over a year ago

@DilipRajkumar What error do you get when you try this solution? This solution is correct way to do what you want, and will not produce the cmd.exe window you wish to avoid (nor will it produce temporary files on your system).

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BullyWiiPlaza · Accepted Answer · 2015-08-25 12:27:19Z

11

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));
}

answered Aug 25, 2015 at 12:27

BullyWiiPlaza

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Comments

Ifesinachi Bryan · Accepted Answer · 2018-09-29 17:38:24Z

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.

answered Sep 29, 2018 at 17:38

Ifesinachi Bryan

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2 Comments

Sebastian Nielsen Over a year ago

Is it synchronous? I.e. does java wait for the exe file to terminate before continuing its execution?

Ifesinachi Bryan Over a year ago

@SebastianNielsen, the ProcessBuilder class is not synchronized. docs.oracle.com/javase/7/docs/api/java/lang/ProcessBuilder.html

Shaikh Arbaaz · Accepted Answer · 2019-12-03 19:41:38Z

5

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 & separator like: &&

edited Dec 3, 2019 at 19:41

answered Jul 30, 2015 at 8:45

Shaikh Arbaaz

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Comments

Deepal Suranga Weerasuriya · Accepted Answer · 2017-05-12 18:11:11Z

0

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"));

answered May 12, 2017 at 18:11

Deepal Suranga Weerasuriya

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