Filename too long

[neil-benn]

When running this as a simpe test; the code is:
// Specify JVM options (optional)
RootExecutor rootExecutor = new RootExecutor("-Xmx64m");

	// Execute privileged action without return
	rootExecutor.run(() -> System.out.println("rmdir \"c:\\Program Files\\test-remove"));

I get an error:

java.io.IOException: Cannot run program ""C:\Users\neilb\AppData\Local\Temp\run-as-root18038586148998856981-elevate"": CreateProcess error=206, The filename or extension is too long

The OS in Win10 64 bit; is there a way to get the filename length down?

[dyorgio]

Hi @neil-benn, tks to open this issue ;).
This link: StackOverflow Post suggests that filename is not the problem, but the classpath.

Take a look in OutProcessUtils.getCurrentClasspath() output (call it in your current project main) and verify if is possible to reduce... you can construct your classpath manually too... Copy source of RootExecutor in a new file and replace OneRunOutProcess constructor call with one that accepts custom classpath.

You maybe will like to use OutProcessUtils.generateClassPath(Class...).

Let me know if you could solve this, I will update library with a new RootExecutor constructor.

[neil-benn]

You are correct; this is a spring boot app so there are a lot of jars in the classpath. It's not so easy to reduce the classpath so something that says 'do not inherit the current classpath would solve the issue. I was just trawling through your source code for outprocess when I came to the same conclusion :).

[neil-benn]

Yep that worked - the actual test doesn't work but the process runs with elevated permissions. Teh new constructor is:

/**
 * Creates an instance with specific java options, the classpath can be specified
 * this is useful if you have a lng classpath that breaks commadn line restrictions on Windows
 *
 * @param classPath a custom classpath
 * @param javaOptions JVM options (ex:"-xmx32m")
 */
public RootExecutor(String classPath, String[] javaOptions) throws IOException {
    this.outProcess = new OneRunOutProcess(MANAGER, classPath, javaOptions);
}

[neil-benn]

However; the fundamental underlying call isn't working too well; I am executing:

	RootExecutor rootExecutor = new RootExecutor(".", new String[] {"-Xmx64m"});
	String command =  "mkdir \"C:\\Program Files\\test-elevate\"";
	System.out.println("executing " + command);
	rootExecutor.run(() -> System.out.println(command));

However; no directory is created. I've obviously checked this in command prompt and it works (as admin) but not in rootexecutor. Am I doing something wrong in the command format?

[neil-benn]

Nope; no matter what I try the command always returns '1' and fails. The elevation works perfectly but the command doesn't.

[dyorgio]

I see two problems:

1. System.out.println will only print argument on process output, not exec command... It is just an example on README :D. Replace with Runtime.exec.
2. "." as classpath is not sufficient, because at least RootExecutor, OutProcess and deps needs to be on classpath.

Use OutProcessUtils.generateClassPath(RootExecutor.class, OneRunOutProcess.class, YOURCLASSTHATISCREATINGLAMBDAORINNERCLASS.class)

[neil-benn]

Thanks; the sysout did seem really weird to me! I see what you mean about the classpath; in fact because of FST it is massive. I note that in the source you have got rid of FST. I think I'll need that version as this makes the dependent libraries much lighter.

[dyorgio]

Done! Version 1.2.3 was released.
Please, note that maybe it take some time to you find it on maven respository.

[dyorgio]

Closed by timeout.