Question

I am using the current expression to convert plain text hyperlink like with http or https to html hyperlinks

String pattern = "(?:https|http)://([^\\s\\|]+)";

However in the message content that will go through this regular expression pattern i have content like http://www.yahoo.com.

For this kind of content in between the two video brackets. I do not want it to be converted to html hyperlink because it will be displayed on the page.

How to resolve this issue?

Answer 1

Use negative lookahead and lookbehind using (?<!...) and (?!...):

String link = "(?:https|http)://([^\\s\\|]+)";
String pattern = "(?<!\\)" + link + "(?!\\)";

Pattern p = Pattern.compile(pattern);
System.out.println(p.matcher("http://www.yahoo.com").matches());
System.out.println(p.matcher("http://www.yahoo.com").matches());

Outputs:

false
true

---

Full example:

public static void main(String... args) {

    Pattern pattern = Pattern.compile(
            "(?<!\\)(?:https|http)://([^\\s\\|]+)(?!\\)");

    String text = "This is some example text with a link" +
        "http://videolink that should not be replaced" + 
        "and another link that should be" +
        "replaced http://www.example.com";

    Matcher m = pattern.matcher(text);

    StringBuffer sb = new StringBuffer();
    while (m.find())
        m.appendReplacement(sb, "<a href=\""+m.group()+"\">"+m.group() +"</a>");
    m.appendTail(sb);

    System.out.println(sb);
}

Outputs:

This is some example text with a link
http://videolink that should not be replaced
and another link that should be
replaced http://www.example.com