Detect and extract url from a string?

Question

This is a easy question,but I just don't get it. I want to detect url in a string and replace them with a shorten one.

I found this expression from stackoverflow,But the result is just http

Pattern p = Pattern.compile("\\b(https?|ftp|file)://[-a-zA-Z0-9+&@#/%?=~_|!:,.;]*[-a-zA-Z0-9+&@#/%=~_|]",Pattern.CASE_INSENSITIVE);         Matcher m = p.matcher(str);         boolean result = m.find();         while (result) {             for (int i = 1; i <= m.groupCount(); i++) {                 String url=m.group(i);                 str = str.replace(url, shorten(url));             }             result = m.find();         }         return html; 

Is there any better idea?

Answer 1

Let me go ahead and preface this by saying that I'm not a huge advocate of regex for complex cases. Trying to write the perfect expression for something like this is very difficult. That said, I do happen to have one for detecting URL's and it's backed by a 350 line unit test case class that passes. Someone started with a simple regex and over the years we've grown the expression and test cases to handle the issues we've found. It's definitely not trivial:

// Pattern for recognizing a URL, based off RFC 3986 private static final Pattern urlPattern = Pattern.compile(         "(?:^|[\\W])((ht|f)tp(s?):\\/\\/|www\\.)"                 + "(([\\w\\-]+\\.){1,}?([\\w\\-.~]+\\/?)*"                 + "[\\p{Alnum}.,%_=?&#\\-+()\\[\\]\\*$~@!:/{};']*)",         Pattern.CASE_INSENSITIVE | Pattern.MULTILINE | Pattern.DOTALL); 

Here's an example of using it:

Matcher matcher = urlPattern.matcher("foo bar http://example.com baz"); while (matcher.find()) {     int matchStart = matcher.start(1);     int matchEnd = matcher.end();     // now you have the offsets of a URL match }