Java : replacing text URL with clickable HTML link

Question

I am trying to do some stuff with replacing String containing some URL to a browser compatible linked URL.

My initial String looks like this :

"hello, i'm some text with an url like http://www.the-url.com/ and I need to have an hypertext link !"

What I want to get is a String looking like :

"hello, i'm some text with an url like <a href="http://www.the-url.com/">http://www.the-url.com/</a> and I need to have an hypertext link !"

I can catch URL with this code line :

String withUrlString = myString.replaceAll(".*://[^<>[:space:]]+[[:alnum:]/]", "<a href=\"null\">HereWasAnURL</a>");

Maybe the regexp expression needs some correction, but it's working fine, need to test in further time.

So the question is how to keep the expression catched by the regexp and just add a what's needed to create the link : catched string

Thanks in advance for your interest and responses !

While the answers below should help you, I suggest you have a look at John Gruber's regular expression to capture urls in all forms that appear "in the wild": daringfireball.net/2009/11/liberal_regex_for_matching_urls — FRotthowe
– FRotthowe, Commented Dec 15, 2009 at 19:06

Paul Croarkin · Accepted Answer · 2010-11-10 17:12:37Z

7

public static String textToHtmlConvertingURLsToLinks(String text) {
    if (text == null) {
        return text;
    }

    String escapedText = HtmlUtils.htmlEscape(text);

    return escapedText.replaceAll("(\\A|\\s)((http|https|ftp|mailto):\\S+)(\\s|\\z)",
        "$1<a href=\"$2\">$2</a>$4");
}

There may be better REGEXs out there, but this does the trick as long as there is white space after the end of the URL or the URL is at the end of the text. This particular implementation also uses org.springframework.web.util.HtmlUtils to escape any other HTML that may have been entered.

answered Nov 10, 2010 at 17:12

Paul Croarkin

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1 Comment

Sonson123 Over a year ago

Doesn't work for two links that are just separated by one space.

Community · Accepted Answer · 2017-05-23 12:09:13Z

6

Try to use:

myString.replaceAll("(.*://[^<>[:space:]]+[[:alnum:]/])", "<a href=\"$1\">HereWasAnURL</a>");

I didn't check your regex.

By using () you can create groups. The $1 indicates the group index. $1 will replace the url.

I asked a simalir question: my question
Some exemples: Capturing Text in a Group in a regular expression

edited May 23, 2017 at 12:09

CommunityBot

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answered Dec 15, 2009 at 18:57

Martijn Courteaux

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1 Comment

Sonson123 Over a year ago

This doesn't work for multiple links in a text because the .* takes too much.

Sonson123 · Accepted Answer · 2015-09-28 11:48:28Z

5

For anybody who is searching a more robust solution I can suggest the Twitter Text Libraries.

Replacing the URLs with this library works like this:

new Autolink().autolink(plainText)

answered Sep 28, 2015 at 11:48

Sonson123

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1 Comment

redochka Over a year ago

url must be well formatted. unable to detect: www.example.com (http:// missing). :(

walkeros · Accepted Answer · 2017-02-14 14:47:50Z

Belows code replaces links starting with "http" or "https", links starting just with "www." and finally replaces also email links.

  Pattern httpLinkPattern = Pattern.compile("(http[s]?)://(www\\.)?([\\S&&[^.@]]+)(\\.[\\S&&[^@]]+)");

  Pattern wwwLinkPattern = Pattern.compile("(?<!http[s]?://)(www\\.+)([\\S&&[^.@]]+)(\\.[\\S&&[^@]]+)");

  Pattern mailAddressPattern = Pattern.compile("[\\S&&[^@]]+@([\\S&&[^.@]]+)(\\.[\\S&&[^@]]+)");

    String textWithHttpLinksEnabled = 
  "ajdhkas www.dasda.pl/asdsad?asd=sd www.absda.pl [email protected] klajdld http://dsds.pl httpsda http://www.onet.pl https://www.onsdas.plad/dasda";

    if (Objects.nonNull(textWithHttpLinksEnabled)) {

      Matcher httpLinksMatcher = httpLinkPattern.matcher(textWithHttpLinksEnabled);
      textWithHttpLinksEnabled = httpLinksMatcher.replaceAll("<a href=\"$0\" target=\"_blank\">$0</a>");

      final Matcher wwwLinksMatcher = wwwLinkPattern.matcher(textWithHttpLinksEnabled);
      textWithHttpLinksEnabled = wwwLinksMatcher.replaceAll("<a href=\"http://$0\" target=\"_blank\">$0</a>");

      final Matcher mailLinksMatcher = mailAddressPattern.matcher(textWithHttpLinksEnabled);
      textWithHttpLinksEnabled = mailLinksMatcher.replaceAll("<a href=\"mailto:$0\">$0</a>");

      System.out.println(textWithHttpLinksEnabled);
    }

Prints:

ajdhkas <a href="http://www.dasda.pl/asdsad?asd=sd" target="_blank">www.dasda.pl/asdsad?asd=sd</a> <a href="http://www.absda.pl" target="_blank">www.absda.pl</a> <a href="mailto:[email protected]">[email protected]</a> klajdld <a href="http://dsds.pl" target="_blank">http://dsds.pl</a> httpsda <a href="http://www.onet.pl" target="_blank">http://www.onet.pl</a> <a href="https://www.onsdas.plad/dasda" target="_blank">https://www.onsdas.plad/dasda</a>

Matt · Accepted Answer · 2009-12-15 19:01:29Z

0

Assuming your regex works to capture the correct info, you can use backreferences in your substitution. See the Java regexp tutorial.

In that case, you'd do

myString.replaceAll(....., "<a href=\"\1\">\1</a>")

answered Dec 15, 2009 at 19:01

Matt

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Comments

alaster · Accepted Answer · 2017-01-06 18:12:44Z

0

In case of multiline text you can use this:

text.replaceAll("(\\s|\\^|\\A)((http|https|ftp|mailto):\\S+)(\\s|\\$|\\z)",
        "$1<a href='$2'>$2</a>$4");

And here is full example of my code where I need to show user's posts with urls in it:

private static final Pattern urlPattern = Pattern.compile(
        "(\\s|\\^|\\A)((http|https|ftp|mailto):\\S+)(\\s|\\$|\\z)");


String userText = ""; // user content from db
String replacedValue = HtmlUtils.htmlEscape(userText);
replacedValue = urlPattern.matcher(replacedValue).replaceAll("$1<a href=\"$2\">$2</a>$4");
replacedValue = StringUtils.replace(replacedValue, "\n", "<br>");
System.out.println(replacedValue);

answered Jan 6, 2017 at 18:12

alaster

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