Java : how to get text between “http://” and first following “/” occurence ? And after first “/” occurence?

Question

I am still a novice with regular expressions, "regex", etc... in Java.

If I have an url like this : " http://somedomain.someextention/somefolder/.../someotherfolder/somepage "

What is the simplest way to get :

• "somedomain.someextention" ?
• "somefolder/.../someotherfolder/somepage" ?
• "somepage" ?

Thanks !

Answer 1

You don't have to (and probably shouldn't) use regex here. Instead use classes defined to handle things like this. You can use for example URL , URI , File classes like

String address = "http://somedomain.someextention/somefolder/.../someotherfolder/somepage";

URL url = new URL(address);
File file = new File(url.getPath());

System.out.println(url.getHost());
System.out.println(url.getPath());
System.out.println(file.getName());

Outpit:

somedomain.someextention
/somefolder/.../someotherfolder/somepage
somepage

Now you can need to get rid of / at start of path to your resource. You can use substring(1) here if resource starts with / .

---

But if you really must use regex you can try with

^https?://([^/]+)/(.*/([^/]+))$

Now

• group 1 will contain host name,
• group 2 will contain path to resource
• group 3 will contain name of resource

Answer 2

The best way to get those components is to use the URI class; eg

    URI uri = new URI(str);
    String domain = uri.getHost();
    String path = uri.getPath();
    int pos = path.lastIndex("/");
    ...
    // or use File to parse the path string.

You could do it using regexes on the raw url string, but there is a risk that you won't correctly cope with all of the variability that is possible in a URL. (Hint: the regex supplied by @Pchenko doesn't :-)) And you would definitely need to use a decoder to deal with possible percent encoding.

Answer 3

This is not a regexp or URI use but simple substring code as an excersise material. Missing few corner case format validation.

int lastDelim = str.lastIndexOf('/);
if (lastDelim<0) throw new IllegalArgumentException("Invalid url");
int startIdx = str.indexOf("//");
startIdx = startIdx<0 ? 0 : startIdx+2;
int pathDelim = str.indexOf('/', startIdx);
String domain = str.substring(startIdx, pathDelim);
String path = str.substring(pathDelim+1, lastDelim);
String page = str.substring(lastDelim+1);

Answer 4

If you would like to use regex to decode the URL instead of using the URI class, as described in the previous answers, the below link gives a nice tutorial of regex, and it explains decoding a sample URL as well. You could learn it there and try it out.

http://www.beedub.com/book/2nd/regexp.doc.html

Answer 5

It's not regex, or scalable at that, it works though:

public class SomeClass
{
    public static void main(String[] args)
    {

        SomeClass sclass = new SomeClass();
        String[] string = 
            sclass.parseURL("http://somedomain.someextention/somefolder/.../someotherfolder/somepage");

        System.out.println(string[0]);
        System.out.println(string[1]);
        System.out.println(string[2]);
    }

    private String[] parseURL(String url)
    {
        String part1 = url.substring("http://".length(), url.indexOf("/", "http://".length()));

        String part2 = url.substring("http://".length() + part1.length() + 1, url.lastIndexOf("/"));

        String part3 = url = url.substring(url.lastIndexOf("/") + 1);

        return new String[] { part1, part2, part3 };
    }
}

Output:

somedomain.someextention
somefolder/.../someotherfolder
somepage