Parsing command line args[], with flags and without using a third party lib

Question

I have a requirement to parse some command line arguments in a Java application without using any third party libraries (like Apache CLI or JCommander). I can only use the standard Java 11 API.

They will be passed into my application on start up and with flags; e.g: java myClass -port 14008 -ip 140.086.12.12 which can be in any order.

I'm pretty new to Java, but how would I go about doing this. Can any standard API functions help with this? Or should I check them manually e.g: if arg[0] = "-port" then arg[0]+1 = port number.

Hi, Input is above, so Java myClass -port 14008 -ip 140.086.12.12 . Output would be 2 variables with the port and ip address in them. — WhatTheWhat
– WhatTheWhat, Commented Mar 25, 2022 at 18:59
"without using any third party libraries" consider why these libraries exist: there would be no need for them if there was a simple solution. Sure, you can do it without, but you will end up with something less functional; you need to decide what functionality you are willing to forgo. — Andy Turner
– Andy Turner, Commented Apr 2, 2022 at 13:00

score 1 · Accepted Answer · 2022-04-02 13:09:38Z

1

The command line will split with white spaces everything you'll write after your "Java MyClass". So, every "word" separated with a space will be treated as a single parameter.

You might wanna pack them together and use a separator character to split them later, like so Java myClass -port:14008 -ip:140.086.12.12.

Or, maybe if you wanna keep your parameter style you need to add the index displacement within brackets, like this

int port = 0;
String ip = "";
for (int i=0; i<args.length; i+=2){
    if (args[i].equals("port")){
       port = Integer.parseInt(args[i+1]);
    } else if (args[i].equals("ip")){
       ip = args[i+1];
    }
}

By the way, this is a very good article that explains quite quickly how Java accpets command line parameters. You might find it useful.

https://www.baeldung.com/java-command-line-arguments

EDIT: thanks to @Andy Turner in the comments

edited Apr 2, 2022 at 13:09

answered Apr 2, 2022 at 12:56

user18680757

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2 Comments

Andy Turner Over a year ago

"Java will split with white spaces" no, the shell will do this.

user18680757 Over a year ago

@AndyTurner you're absolutely right. I'll edit it!

WhatTheWhat · Accepted Answer · 2022-04-21 09:36:18Z

This is how I solved the problem. It assumes there are class member variables set with default values to overide.

private String ipAddress = "localhost";

private Integer port = 14001;

     /**
     * Method to get the ip and port number from the command line args.
     * It caters for any order of flag entry.It tries to override the
     * defaults set as a class variables, so if it can't, it uses those.
     *
     * @param args The args passed in by the user.
     */
    private void getPortAndIp(String[] args) {
        System.out.println("Getting inputs ");
        if ((args[0].equals("-ccp")) && (args.length == 2)) {
            this.port = Integer.parseInt(args[1]);
        } else if ((args[0].equals("-cca")) && (args.length == 2)) {
            this.ipAddress = args[1];
        } else if ((args[0].equals("-ccp")) && (args[2].equals("-cca"))
                && (args.length == 4)) {
            this.port = Integer.parseInt(args[1]);
            this.ipAddress = args[3];
        } else if ((args[0].equals("-cca")) && (args[2].equals("-ccp"))
                && (args.length == 4)) {
            this.port = Integer.parseInt(args[3]);
            this.ipAddress = args[1];
        } else {
            System.out.println("Options:");
            System.out.println("-ccp [port number]");
            System.out.println("-cca [ip address]");
            System.out.println("Could not determine port from command line " +
                    "arguments, using defaults: ");
        }
        System.out.println("on " + this.ipAddress + ":" + this.port);
    }

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