Parse URL Hash with Regex

Question

Need to parse this String

#Login&oauth_token=theOAUTHtoken&oauth_verifier=12345

Where I just need to get oauth_token and oauth_verifier key + values, what is the simplest way to do this with Regex?

With all the rep you have, you should know better than to ask a question without code or effort shown... — Jerry, Aug 18 '13 at 17:21
Does this help:- http://stackoverflow.com/questions/27745/getting-parts-of-a-url-regex ??? — Rahul Tripathi, Aug 18 '13 at 17:24

hwnd · Accepted Answer · 2013-08-18T18:12:39.007

This will do it, you did not specify how you wanted your data output so I seperated them with a comma.

import java.util.regex.*;

class rTest {
  public static void main (String[] args) {
    String in = "#Login&oauth_token=theOAUTHtoken&oauth_verifier=12345";
    Pattern p = Pattern.compile("(?:&([^=]*)=([^&]*))");
    Matcher m = p.matcher(in);
    while (m.find()) {
      System.out.println(m.group(1) + ", " + m.group(2));
    }
  }
}

Regular expression:

(?:           group, but do not capture:
  &           match '&'
   (          group and capture to \1:
    [^=]*     any character except: '=' (0 or more times)
   )          end of \1
   =          match '='
   (          group and capture to \2:
    [^&]*     any character except: '&' (0 or more times)
   )          end of \2
)             end of grouping

Output:

oauth_token, theOAUTHtoken
oauth_verifier, 12345

score 0 · Answer 2 · answered Aug 18 '13 at 17:26

This should work:

String s = "#Login&oauth_token=theOAUTHtoken&oauth_verifier=12345";
Pattern p = Pattern.compile("&([^=]+)=([^&]+)");
Matcher m = p.matcher(s.substring(1));
Map<String, String> matches = new HashMap<String, String>();
while (m.find()) {
    matches.put(m.group(1), m.group(2));
}
System.out.println("Matches => " + matches);

OUTPUT:

Matches => {oauth_token=theOAUTHtoken, oauth_verifier=12345}

Parse URL Hash with Regex

2 Answers2