How to split URL and take only the host names

Question

Im trying to split URLs, for example https://stackoverflow.com/questions/ and take only stackoverflow.com. How can I do this in Java without using the built in function getHost()?

See https://stackoverflow.com/questions/4826061/what-is-the-fastest-way-to-get-the-domain-host-name-from-a-url — Bert Maurau, Oct 31 '19 at 23:46
Why not use the tools of the language? Why reinvent a URL parser when you already have one, that has been fully tested, and is known to work? — Andreas, Oct 31 '19 at 23:55

Jason · Answer 1 · 2019-11-01T00:18:17.220

0

One thing you can do is use String#replaceAll. I know it's not what you want but off the bat it's a decent way to do it.

String uri = "https://stackoverflow.com/questions/";

if (uri.contains("https://")) {
    uri = uri.replaceAll("https://", "");
}

if (uri.contains("http://")) {
    uri = uri.replaceAll("http://", "");
}

int indexOfBackwardsSlash = uri.indexOf("/");

if (indexOfBackwardsSlash != -1) {
    uri = uri.substring(0, indexOfBackwardsSlash);
}

Use URI#getPath.

URI uri = URI.create("https...");

String path = uri.getPath();

edited Nov 01 '19 at 00:18

answered Oct 31 '19 at 23:46

Jason

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The getPath() returns the path after the host - you means getHost more likely ? – Abs Nov 01 '19 at 00:53
Actually getPath returns the rawPath, which is; "The string returned by this method is equal to that returned by the getRawPath method except that all sequences of escaped octets are decoded.". – Jason Nov 01 '19 at 01:04

score 0 · Answer 2 · answered Nov 01 '19 at 00:08

If you can put your URL into a String , there is this option :

public static void main(String []args){
  String str ="https://stackoverflow.com/questions/";
  String[] parts = str.split("/");
  String part1 = parts[0]; //https:
  String part2 = parts[1]; //'nothing'
  String part3 = parts[2]; //stackoverflow.com
  String part4 = parts[3]; //questions
}

score 0 · Answer 3 · answered Nov 01 '19 at 00:35

You could also use regular expressions:

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class UrlRegex {
     public static void main(String []args){
        String url = "https://stackoverflow.com/questions/";
        Pattern pat = Pattern.compile("//([^/]*)"); //match everything between "//" and "/"
        Matcher m = pat.matcher(url);
        if (m.find()) {
            String hostname = m.group(1);
        }
     }
}

Won't match stackoverflow.com/questions/ ! – Abs Nov 01 '19 at 01:02 — Abs, Nov 01 '19 at 01:02

score 0 · Answer 4 · answered Nov 01 '19 at 00:51

Here you go :

    Pattern pattern = Pattern.compile("^(?:(?:http)s?://)?(?<hostGroup>[^/:]+).*");
    Matcher matcher = pattern.matcher("https://stackoverflow.com/questions/");

    if (matcher.matches()) {
        System.out.println(matcher.group("hostGroup"));
    } else {
        System.out.println("Not found! Invalid URL ^^");
    }

The above will find stackoverflow.com for the following urls strings :

https://stackoverflow.com/questions/
http://stackoverflow.com/questions/
stackoverflow.com/questions/
stackoverflow.com:80/questions/
stackoverflow.com/
stackoverflow.com

I guess that's for practicing regex ? Otherwise, prefer using the standard APIs whenever possible - (in your case URI#getHost() !)

Cheers!

score 0 · Answer 5 · edited Nov 10 '19 at 10:03

If you are sure that you are getting the proper URL format than you can just substring it preferred places.

    public static void main(String[] args) {
        String url = "https://stackoverflow.com/questions/";
        System.out.println(getHostFast(url));
    }
    public static String getHostFast(String url) {
        String subbed = url.substring(url.indexOf('/') + 2);
        return subbed.substring(0, subbed.indexOf('/'));
    }

The error prof method would need to contain additional check (for example if the next '/' exists after dropping http://.

How to split URL and take only the host names

5 Answers5