5

What do you have to do in Java to get the Runtime.exec() to run a program that is on the path? I'm trying to run gpsbabel which I have put into the path (/usr/local/bin).

public class GpxLib {

    public static void main(String[] args) {
        try
        {
            Runtime r = Runtime.getRuntime();
            Process p = r.exec("gpsbabel -i garmin -f usb: -o gpx -F -");
            InputStream is = p.getInputStream();
            BufferedReader br = new BufferedReader(new InputStreamReader(is));
            while (true)
            {
                String s = br.readLine();
                if (s == null)
                    break;
                System.out.println(s);
            }
            br.readLine();
        } catch (IOException e) {
            e.printStackTrace(System.err);
        }
    }
}
Jason S
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3 Answers3

4

I added a call to System.out.println(System.getenv("PATH")); which only prints out

/usr/bin:/bin:/usr/sbin:/sbin

so for some reason /usr/local/bin doesn't show up. Looks like this is a MacOSX question or an Eclipse question, not a Java question. edit: asked this question on superuser instead.

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Jason S
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3

It will inherit the path from the Java process. So whatever environment the Java process has, the spawned process will have as well. Here's how to check the environment:

Map<String, String> env = System.getenv();
for (String envName : env.keySet()) {
     System.out.format("%s=%s%n", envName, env.get(envName));
}

Have you set the PATH and exported it ? If you don't export it, then it's not available to subprocesses.

Additionally, you must consume stdout and stderr concurrently, to prevent blocking. Otherwise stuff will work in some circumstances, then your spawned process will output a different quantity of data and everything will grind to a halt.

See this answer for more details.

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Brian Agnew
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  • I haven't set the path at all, anywhere. It is whatever the default path is for Eclipse and for bash. – Jason S Aug 23 '09 at 19:24
  • The path will be set per user. You may want to add the path entry for that executable in your .bashrc (and perhaps login again if you're launching Eclipse from the toolbar or similar) – Brian Agnew Aug 23 '09 at 19:53
  • I think the *real* solution is to specify the complete path to the binary, however, so that your program isn't environment dependent, and then make that a configuration item. That's usually my *strategic* approach. – Brian Agnew Aug 23 '09 at 19:55
  • agreed, I suppose the better way is to use a .properties file.... I just hate having to run java programs like "java -Dsome.ugly.variable=/usr/local/bin/gpsbabel -jar myjar.jar" – Jason S Aug 23 '09 at 19:58
  • ...and I just want to put gpsbabel someplace appropriate so I can access it on the path. I asked a question somewhere (dammit, can't find which question, i thought it was on superuser or serverfault) and /usr/local/bin was the recommended place. – Jason S Aug 23 '09 at 20:01
2

Here is the solution:

ProcessBuilder proc = new ProcessBuilder("<Directory PAth>" + "Executable.exe");
proc.redirectOutput(ProcessBuilder.Redirect.INHERIT);
proc.directory(fi); //fi = the output directory path
proc.start();

is the path where program\application's excutable is located e.g "C:\MyProg\"

DareDevil
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  • anybody can explain what is wrong with this code? I have Tested successfully by calling a C++ console application in it. – DareDevil Dec 09 '13 at 12:39