Regex to extract a substring

[Chandra shekar M]

Hi All

In the below Regex code

public static void main(String... s){
String url="http://[e-mail]:8080";
		Pattern pattern = Pattern.compile("(.*://)?(.*):([0-9]+)");
		Matcher matcher = pattern.matcher(url);
		while(matcher.find()){
			System.out.println(" tuned "+matcher.start()+"  "+matcher.end()+" "+matcher.group()
				+"  "+matcher.group(1) +"  "+matcher.group(2)+ " "+matcher.group(3)
				+" count "+ matcher.groupCount());
			
		}
		}

I get three groups http:// , [e-mail] and 8080. So in this case I am matching all the three groups. Is there a way where i can directly extract only last two groups without matching the first group. say like this Pattern.compile((.*) : ([0-9]+)). but the problem with this is (.*) will also match the http:// and i'll get http://[e-mail] and 8080. So i want to define a regex in such a way that i should be able to extract everything after http:// and out put should be [e-mail] and 8080.
How can i do this?

Thanks
Chandra

[Winston Gutkowski]

Chandra shekar M wrote:In the below Regex code

Chandra,

I've split up that enormous line in your code. Please re-read the UseCodeTags (←click) page thoroughly.

Winston

[Winston Gutkowski]

Chandra shekar M wrote:So i want to define a regex in such a way that i should be able to extract everything after http:// and out put should be [e-mail] and 8080.
How can i do this?

Just remove the brackets from the first expression. However, I wonder why you think you need to do this. With 3 groups you can still get the last two, so why bother?

One point: you may make things quite a bit faster by using the possessive qualifier and more specific search criteria, viz:
Pattern.compile("([^:]+://)?+([^:]*+):([0-9]+)")

Winston

[Stuart A. Burkett]

Or you could use the URL class and use its various getXXX methods to extract the parts you want.