Formatting IP:Port string to

Question

I\'m trying to make a little chat program for experimentation, but seeing as I\'m not the best Java programmer, I don\'t know how to separate a port from an IP where they are bo

Answer 1

First, you should check if the String contains a colon. Then, you can use String.split(String) and Integer.parseInt(String) with something like

String input = "127.0.0.1:8080"; // <-- an example input
int port = 80; // <-- a default port.
String host = null;
if (input.indexOf(':') > -1) { // <-- does it contain ":"?
  String[] arr = input.split(":");
  host = arr[0];
  try {
    port = Integer.parseInt(arr[1]);
  } catch (NumberFormatException e) {
    e.printStackTrace();
  }
} else {
  host = input;
}
System.out.printf("host = %s, port = %d%n", host, port);

Output is

host = 127.0.0.1, port = 8080

Answer 2

Use String.split
IP = splittedIP[0], Port = splittedIP[1] in String
you will need Integer.parseInt to get the integer value of port

String[] ipSplit = "127.0.0.1:80".split(":");
String ip = ipSplit[0];
int port = Integer.parseInt(ipSplit[1]);

Answer 3

String s = "127.0.0.1:999";
String[] parts = s.split(":");
String address = parts[0];
int port = Integer.parseInt(parts[1]);

In this case, address will be "127.0.0.1" and port will be the int 999.

Note that Integer.parseInt will throw a NumberFormatException in this case if the portion of the string after the : cannot be parsed as an integer, as in "127.0.0.1:blahblah". As well, if the string does not contain a : there will be no parts[1].

Answer 4

public static void main(String args[]){
        String allTogether= "ip:port";
        String[] array;

        if(allTogether.contains(":")){
            array = allTogether.split(":");
            String ip = array[0];
            String port = array[1];

            System.out.println(ip);
            System.out.println(port);
        }
    }

Answer 5

to validate IP:PORT format you may try this:

public class Validator {

    private static final String IP_PORT_PATTERN =
            "^([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." +
                    "([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." +
                    "([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." +
                    "([01]?\\d\\d?|2[0-4]\\d|25[0-5]):" +
                    "([1-9][0-9]{0,3}|[1-5][0-9]{4}|6[0-4][0-9]{3}|65[0-4][0-9]{2}|655[0-2][0-9]|6553[0-5])$";

    private static final Pattern PATTERN;

    static {
        PATTERN = Pattern.compile(IP_PORT_PATTERN);
    }

    public static boolean validate(final String s) {
        return PATTERN.matcher(s).matches();
    }

}

thanks to frb and mkyong :)

and to separate ip and port :

public static void main(String args[]) {
    String s = "127.0.0.1:8080";
    if (validate(s)) {
        String ip = s.substring(0, s.indexOf(':'));
        int port = Integer.parseInt(s.substring(s.indexOf(':') + 1));
        System.out.println(ip);
        System.out.println(port);
    } else {
        System.out.println("invalid format");
    }
}

Answer 6

You can use pattern matcher in java to get the address and the port information and by the way you can validate the input string as well.

Pattern pattern = Pattern.compile(REGEX_FOR_IPADDRESS_WITH_PORT);
Matcher matcher = pattern.matcher("127.0.0.1:80");
if (matcher.matches()) {
    System.out.printf("host = %s, port = %s%n", matcher.group(1), matcher.group(2));
} else {
    System.out.println("Invalid Ip Address");
}

You can have multiple regular expression to validate V4 and V6 addresses. Hope this would help.