Java Programming

Try this

String cmd = "java -jar \"" ++workingDir ++"\"/SwingApplication.jar"; 

Edited by: DynamicBasics on Jul 15, 2010 5:31 PM

A good case for using ProcessBuilder.

String cmd = "java -jar \"" ++workingDir ++"\"/SwingApplication.jar";

doesnt seem to work, so i tried using process builder as suggested

ProcessBuilder pb = new ProcessBuilder("java -jar");
pb.directory(new File(workingDirPath + "/SwingApplication.jar"));
Process p = pb.start();

. but getting an exception

java.io.IOException: Cannot run program "java -jar" (in directory "C:\Run Application\SwingApplication.jar"): CreateProcess error=267, The directory name is invalid

Do I need to change the directory path in this?

While..

C:\Run Application\

..might be a directory, is..

C:\Run Application\SwingApplication.jar

?

I was thinking something more like this (untested)..

String path = workingDirPath + "/SwingApplication.jar";
String[] command = {
	"java", 
	"-jar",
	path
};
ProcessBuilder pb = new ProcessBuilder(command);
Process p = pb.start();

BTW - when posting code, code snippets, HTML/XML or input/output, please use the code tags. The code tags protect the indentation anf formatting of the sample. To use the code tags, select the sample and click the CODE button.

The command to be finally executed is

C:\Program Files\Java\jdk1.5.0\bin\java.exe -jar C:\Run Application\SwingApplication.jar

as the path has spaces, enclose in double quotes

"C:\Program Files\Java\jdk1.5.0\bin\java.exe" -jar "C:\Run Application\SwingApplication.jar"

This goes into the string variable like this

String cmd = "\"C:\\Program Files\\Java\\jdk1.5.0\\bin\\java.exe\" -jar \"C:\\Run Application\\SwingApplication.jar\"";

check the above command once.

Without the absolute path to the JAVA file, the control probably is searching for it in the current working directory.

thank you so much it worked now.

You are welcome, and it is a good practice to mark the question as 'answered'.

Edited by: DynamicBasics on Jul 15, 2010 10:38 PM