CODEWITHSUNDEEP
javaregexassertionregex-lookaroundslookbehind
Nahydrin
Nahydrin

Reputation: 13507

Regex capture lookbehind and lookahead

I'm trying to write regex for the following situations:

badword%
%badword
%badword%

The % signs differ, depending on where they are. A % at the front needs a lookbehind to match letters preceding the word badword until it reaches a non-letter. Likewise, any % that is not at the front needs a lookahead to match letters following the word badword until it hits a non-letter.

Here's what I'm trying to achieve. If I have the following:

Just a regular superbadwording sentece.

badword   # should match "badword", easy enough
badword%  # should match "badwording"
%badword% # should match "superbadwording"

At the same time. If I have a similar sentence:

Here's another verybadword example.

badword   # should match "badword", easy enough
badword%  # should also match "badword"
%badword% # should match "verybadword"

I don't want to use spaces as the assertion capture groups. Assume that I want to capture \w.

Here's what I have so far, in Java:

String badword  = "%badword%";
String _badword = badword.replace("%", "");
badword = badword.replaceAll("^(?!%)%", "(?=\w)"); // match a % NOT at the beginning of a string, replace with look ahead that captures \w, not working
badword = badword.replaceAll("^%", "(?!=\w)"); // match a % at the beginning of a string, replace it with a look behind that captures \w, not working
System.out.println(badword); // ????

So, how can I accomplish this?

PS: Please don't assume the %'s are forced to the start and end of a match. If a % is the first character, then it will need a look behind, any and all other %'s are look aheads.

Upvotes: 3

Views: 5947

Answers (2)

Steve P.
Steve P.

Reputation: 14699

badword = badword.replaceAll("^%", "(?!=\w)"); 
// match a % at the beginning of a string, replace it with a look behind 
//that captures \w, not working

(?!=\w) is a negative-look ahead for =\w, but it seems like you want a positive look-behind. Secondly, lookaheads and lookbehinds are atomic, and thus inherently not capturing, so if I'm correct in my interpretation, you want:

"(?<=(\\w+))". You need the additional () for capturing. For your first part, it would be: "(?=(\\w+)), and the first argument should be "(?<!^)%".

PS: You need two backslashes for \\w, and you seem to want to match multiple characters, no? If so, you would need \\w+. Also, if you don't want to do this for every occurrence, then I suggest using String.format() instead of replaceAll().

Upvotes: 3

LeonidasCZ
LeonidasCZ

Reputation: 153

From your question it doesn't seem necessary to use lookaround, so you could just replace all % with \w*

Snippet:

String tested = "Just a regular superbadwording sentece.";
String bad = "%badword%";
bad = bad.replaceAll("%", "\\\\w*");
Pattern p = Pattern.compile(bad);
Matcher m = p.matcher(tested);
while(m.find()) {
    String found = m.group();
    System.out.println(found);
}

\w doesn't match #,-,etc. so I think \S is better here

Upvotes: 2

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