extract substring in java using regex

Question

I need to extract \"URPlus1_S2_3\" from the string:

\"Last one: http://abc.imp/Basic2#URPlus1_S2_3,\" 

using regular expressi

Answer 1

String s = "Last one: http://abc.imp/Basic2#URPlus1_S2_3,";
Matcher m = Pattern.compile("(URPlus1_S2_3)").matcher(s);
if (m.find()) System.out.println(m.group(1));

You gotta learn how to specify your requirements ;)

Answer 2

String s = Last one: http://abc.imp/Basic2#URPlus1_S2_3,"
String result = s.replaceAll(".*#", "");

The above returns the full String in case there's no "#". There are better ways using regex, but the best solution here is using no regex. There are classes URL and URI doing the job.

Answer 3

Try

Pattern p = Pattern.compile("#([^,]*)");
Matcher m = p.matcher(myString);
if (m.find()) {
  doSomethingWith(m.group(1));  // The matched substring
}

Answer 4

You haven't really defined what criteria you need to use to find that string, but here is one way to approach based on '#' separator. You can adjust the regex as necessary.

expr: .*#([^,]*)
extract: \1

Go here for syntax documentation:

http://download.oracle.com/javase/1.4.2/docs/api/java/util/regex/Pattern.html

Answer 5

Here's a long version:

String url = "http://abc.imp/Basic2#URPlus1_S2_3,";
String anchor = null;
String ps = "#(.+),";
Pattern p = Pattern.compile(ps);
Matcher m = p.matcher(url);
if (m.matches()) {
    anchor = m.group(1);
}

The main point to understand is the use of the parenthesis, they are used to create groups which can be extracted from a pattern. In the Matcher object, the group method will return them in order starting at index 1, while the full match is returned by the index 0.

Answer 6

If you just want everything after the #, use split:

String s = "Last one: http://abc.imp/Basic2#URPlus1_S2_3," ;
System.out.println(s.split("#")[1]);

Alternatively, if you want to parse the URI and get the fragment component you can do:

URI u = new URI("http://abc.imp/Basic2#URPlus1_S2_3,");
System.out.println(u.getFragment());