Detect and extract url from a string?

Asked 2011-04-19T08:24:45.033

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?

java regex url

Edit Report

3 Answers

105

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
}

answered 2011-04-19T08:53:08.993

m.group(1) gives you the first matching group, that is to say the first capturing parenthesis. Here it's (https?|ftp|file)

You should try to see if there is something in m.group(0), or surround all your pattern with parenthesis and use m.group(1) again.

You need to repeat your find function to match the next one and use the new group array.

answered 2011-04-19T08:30:58.813

Detecting URLs is not an easy task. If its enough for you to get a string that starts with https?|ftp|file then it could be fine. Your problem here is, that you have a capturing group, the () and those are only around the first part http...

I would make this part a non capturing group using (?:) and put brackets around the whole thing.

"\\b((?:https?|ftp|file)://[-a-zA-Z0-9+&@#/%?=~_|!:,.;]*[-a-zA-Z0-9+&@#/%=~_|])"

answered 2011-04-19T08:37:28.257

Detect and extract url from a string?

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