将php regex转换为android java regex

Question

Can somebody help me with converting a PHP regex to Java format?

It would be great and I would appreciate if anyone could help me, because I'm not so strong in regex.

This is a PHP code:

$string = file_get_contents('a.txt');
$regex = '/\b(http|https):\/\/[-A-Z0-9+&@#\/%?=~_|$!:,.;]*[A-Z0-9+&@#\/%=~_|$]/i';
preg_match_all($regex, $string, $matches);
$urls = $matches[0];

Answer 1

You should double escape the back slashes and you do not have to escape / s:

String rx = "(?i)\\b(http|rtmp|rtsp|mmsh|mms)://[-A-Z0-9+&@#/%?=~_|$!:,.;]*[A-Z0-9+&@#/%=~_|$]";

Sample code to get all matches (do not forget about including import java.util.regex.*; ):

String str = "http://google.com";

String rx = "(?i)\\b(http|rtmp|rtsp|mmsh|mms)://[-A-Z0-9+&@#/%?=~_|$!:,.;]*[A-Z0-9+&@#/%=~_|$]";
Pattern ptrn = Pattern.compile(rx);
Matcher m = ptrn.matcher(str);
while (m.find()) {
   System.out.println(m.group(0));
}

Answer 2

Basic Regex syntax are same in all language Just be careful around the escape character things. To escape special characters use \\\\ , there is no need to escape /

and some parameter will be given in a different way in java like.

use (?i) at the beginning instead of /i at the end

boundary expression \\b also same in java but you will need to add \\\\b in java because \\b will be considered as a backspace back-space character

Your pattern will look like:

(?i)\\\\b(http|rtmp|rtsp|mmsh|mms)://[-A-Z0-9+&@#/%?=~_|$!:,.;]*[A-Z0-9+&@#/%=~_|$]