Java - Regular Expressions matching one to another

Question

I am trying to retrieve bits of data using RE. Problem is I'm not very fluent with RE. Consider the code.

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

class HTTP{

    private static String getServer(httpresp){
        Pattern p = Pattern.compile("(\bServer)(.*[Server:-\r\n]"); //What RE syntax do I use here?
        Matcher m = p.matcher(httpresp);

        if (m.find()){
            return m.group(2);

    public static void main(String[] args){
        String testdata = "HTTP/1.1 302 Found\r\nServer: Apache\r\n\r\n"; //Test data

        System.out.println(getServer(testdata));

How would I get "Server:" to the next "\\r\\n" out which would output "Apache"? I googled around and tried myself, but have failed.

Answer 1

It's a one liner:

private static String getServer(httpresp) {
    return httpresp.replaceAll(".*Server: (.*?)\r\n.*", "$1");
}

The trick here is two-part:

• use .*? , which is a reluctant match (consumes as little as possible and still match)
• regex matches whole input, but desired target captured and returned using a back reference

Answer 2

You could use capturing groups or positive lookbehind.

Pattern.compile("(?:\\bServer:\\s*)(.*?)(?=[\r\n]+)");

Then print the group index 1.

Example:

String testdata = "HTTP/1.1 302 Found\r\nServer: Apache\r\n\r\n";
Matcher matcher = Pattern.compile("(?:\\bServer:\\s*)(.*?)(?=[\r\n]+)").matcher(testdata);
if (matcher.find())
{
    System.out.println(matcher.group(1));
}

OR

Matcher matcher = Pattern.compile("(?:\\bServer\\b\\S*\\s+)(.*?)(?=[\r\n]+)").matcher(testdata);
if (matcher.find())
{
    System.out.println(matcher.group(1));
}

Output:

Apache

Explanation:

• (?:\\\\bServer:\\\\s*) In regex, non-capturing group would be represented as (?:...) , which will do matching only. \\b called word boundary which matches between a word character and a non-word character. Server: matches the string Server: and the following zero or more spaces would be matched by \\s*

• (.*?) In regex (..) called capturing group which captures those characters which are matched by the pattern present inside the capturing group. In our case (.*?) will capture all the characters non-greedily upto,

• (?=[\\r\\n]+) one or more line breaks are detected. (?=...) called positive lookahead which asserts that the match must be followed by the characters which are matched by the pattern present inside the lookahead.