如何使用执行选项创建Jar文件？

Question

How can I create a jar file with execution options, for example, the idea is execute the command:

java -jar MyProgram.jar -M someFile.txt

or

java -jar MyProgram.jar -cp someFile.txt

So the -M option defines a particular method to process the file someFile.txt and with -cp defines another behavior for the code.

With this in mine, how can I get this result from my code, is there something I need to write in the Main class, or how can I define such behavior?

Answer 1

I think you probably need to check the Apache Commons-CLI, it allows you to do the things you describe above, also I give an example how it works, it give the way to specify a message for the usage of the arguments:

https://commons.apache.org/proper/commons-cli/introduction.html

 Options options = new Options();

    options.addOption( "M", false,"Merge files request.")
            .addOption("CP", false,"Copy files from file.");

    CommandLineParser parser = new DefaultParser();
    try {

        CommandLine cmd = parser.parse(options, args);
        if (!cmd.hasOption("M")) {
            throw new IllegalArgumentException("Must specify an input file.");
        }
        // Do something

        if (!cmd.hasOption("CP")) {
            throw new IllegalArgumentException("Must specify an input file.");
        }

       // Do something

    catch (Exception e) {
        System.out.println(e.getMEssage());
    }

Answer 2

take a look at this example .

basically the args in your main method

public static void main(String[] args) { ... }

args = the arg ument s you put after java -jar MyJar.jar for example -cp someFile.txt as a String[]: {"-cp", "someFile.txt"}