'Regex for matching something if it is not preceded by something else

With regex in Java, I want to write a regex that will match if and only if the pattern is not preceded by certain characters. For example:

String s = "foobar barbar beachbar crowbar bar ";

I want to match if bar is not preceded by foo. So output would be:

barbar
beachbar
crowbar
bar
javaregex


Solution 1:[1]

You want to use negative lookbehind like this: 

\w*(?<!foo)bar

Where (?<!x) means "only if it doesn't have "x" before this point".

See Regular Expressions - Lookaround for more information.

Edit: added the \w* to capture the characters before (e.g. "beach").

Solution 2:[2]

Another option is to first match optional word characters followed by bar, and when that has matched check what is directly to the left is not foobar.

The lookbehind assertion will run after matching bar first.

\w*bar(?<!foobar)
  • \w* Match 0+ word characters
  • bar Match literally
  • (?<!foobar) Negative lookbehind, assert from the current position foobar is not directly to the left.

Regex demo

Solution 3:[3]

In some cases, it could be easier to optionally include the preceding part, then skip those matches in a second step. For instance, to find numbers that don't start with a "+":

if (preg_match_all('/(\+?[0-9][0-9\s\-].*[0-9])/s',$text,$matches)) {
    foreach($matches[1] as $match) {
        if(substr($match,0,1) == '+'){
            continue;
        }
        // continue processing
    }
}

The negative look behind did not work since it would still match 2+ digits, but it would not include the first digit in the match. For instance +1234 would be returned as 234.

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 de-russification
Solution 2
Solution 3 Frank Forte

Related Questions