Why am I getting IOException : "Cannot run program ./shellScript.sh error = 2, No such file or directory in Java?

Question

The following Main.java code simply tries to simulate the following linux command:

cd /dir1/dir2
./shellScript.sh

The program below works only if the executable Main.jar sits within /dir1/dir2, not outside of /dir1/dir2. How do I modify the program below so that Main.jar can sit anywhere on the file system?

public class Main {

    public static String runCmdLineProcess(String commandStr){
        String returnVal = "";
        Runtime r = Runtime.getRuntime();
        
        try {
            Process p = r.exec(commandStr);
            BufferedReader b = new BufferedReader(new InputStreamReader(p.getInputStream()));
            String line = ""; 
            
            while ((line = b.readLine()) != null){
                returnVal += line + "\n";
            }

        }
        catch(IOException ex){
            ex.printStackTrace();
        }
        
        return returnVal;
    }

    public static void runProcessBuilder(String scriptPath){

        String[] cmd = {scriptPath};
        try {
            runCmdLineProcess("cd /dir1/dir2");
            Runtime.getRuntime().exec(cmd);
        }
        catch (IOException ex){
            ex.printStackTrace();
        }
    }
    
    public static void main(String[] args){
        runProcessBuilder("./shellScript.sh"); // <-- works if I run from inside "/dir1/dir2".  
                                               //But if I'm outside "dir2", get an error message 
                                               // saying "Cannot run program "./shellScript.sh": error = 2, No such file or directory
    }

}

There is an overloaded exec that lets you specify the working directory. — f1sh
– f1sh, Commented Mar 15, 2022 at 19:27
Likely issues: (1) you're running two processes, so 'cd' in one cannot affect the other, (2) 'cd' is not a separate executable anyway, it's a shell built-in. But as the other comment says, it's the wrong way to go about it anyway. — passer-by
– passer-by, Commented Mar 15, 2022 at 19:48
String[] cmd = { "/bin/bash" , "-c", "cd /zookeeper/bin" + "./zkCli.sh"}; Runtime run = Runtime.getRuntime(); Process pr = run.exec(cmd); pr.waitFor(); BufferedReader buf = new BufferedReader(new InputStreamReader(pr.getInputStream())); String line = ""; while ((line=buf.readLine())!=null) { System.out.println(line); } whats wrog in this — aΨVaN
– aΨVaN, Commented May 25, 2022 at 18:19

DuncG · Accepted Answer · 2022-03-15 19:50:39Z

1

You should use ProcessBuilder to launch or one of the overloads of exec. You need to specify the pathname to the script and pass the same pathname as the current directory to run the script in:

File pwd = new File("/dir1/dir2");
String shell = new File(pwd, "shellScript.sh").toString();

ProcessBuilder pb = new ProcessBuilder(shell);
// No STDERR => merge to STDOUT - or call redirectError(File)
pb.redirectErrorStream(true);
// Set CWD for the script
pb.directory(pwd);

Process p = pb.start();

// Move STDOUT to the output stream (or original code to save as String)
try(var stdo = p.getInputStream()) {
    stdo.transferTo(stdout);
}

int rc = p.waitFor();

answered Mar 15, 2022 at 19:50

DuncG

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