Regex for matching something if it is not preceded by something else
JavaRegexJava Problem Overview
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
Java Solutions
Solution 1 - Java
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 - Java
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 charactersbar
Match literally(?<!foobar)
Negative lookbehind, assert from the current positionfoobar
is not directly to the left.
Solution 3 - Java
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.