I need to extract "URPlus1_S2_3" from the string:
"Last one: http://abc.imp/Basic2#URPlus1_S2_3,"
using regular expression in Java language.
Can someone please help me? I am using regex for the first time.
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2Please add some more criteria, if you only have to extract that string then you might as well copy it. Is your criterium that you have to find the part between '#' and ',' ?Sebastiaan van den Broek– Sebastiaan van den Broek2011-01-20 16:16:17 +00:00Commented Jan 20, 2011 at 16:16
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extracting between '#' and ',' seem to be the most logical approach to me.Niras– Niras2011-01-20 16:29:21 +00:00Commented Jan 20, 2011 at 16:29
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6 Answers
Try
Pattern p = Pattern.compile("#([^,]*)");
Matcher m = p.matcher(myString);
if (m.find()) {
doSomethingWith(m.group(1)); // The matched substring
}
Comments
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
answered Jan 20, 2011 at 16:15
Konstantin Komissarchik
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3 Comments
Christoffer Hammarström
You missed the trailing comma.
Mike Samuel
This also matches the comma which he did not want matched, and will not match if there are newlines between the hash and comma.Excluding those was not specified.
Konstantin Komissarchik
Indeed, regarding comma. Fixed.
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.
Comments
Since it's the first time you use regular expressions I would suggest going another way, which is more understandable for now (until you master regular expressions ;) and it will be easily modified if you will ever need to:
String yourPart = new String().split("#")[1];
answered Jan 20, 2011 at 16:17
Alberto Zaccagni
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Comments
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.
2 Comments
Niras
This couldn't' get me the answer I needed.I think the String ps is not correct.
Nicolas Buduroi
Well, it should! I'm using Clojure, but this language is using Java's regex engine directly:
(second (re-find #"#(.+)," "Last one: http://abc.imp/Basic2#URPlus1_S2_3,")) results in "URPlus1_S2_3".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());
