I tried \b (This means the last character of a word) in the Java Regexp, but this doesn't work.
String input = "aaa aaa";
Pattern pattern = Pattern.compile("(a\b)");
Matcher matcher = pattern.matcher(input);
while (matcher.find()) {
System.out.println("Found this wiki word: " + matcher.group());
}
What in the problem?
In Java, "\b" is a back-space character (char 0x08), which when used in a regex will match a back-space literal.
You want the regex a\b, which in java is coded by escaping the back-slash, like this:
"a\\b"
btw, you are only partially correct about the meaning of regex \b - it actually means "word boundary" (either the start or end of a word).
\b matches in unexpected places (visualized by |: Ü|bertr|ä|ger|).Literal backslashes in Java strings need escaping, so the regex \b becomes "\\b" as a Java string.
\` and \b` are escape sequence (character) in Java.
\bdoes not mean "last character of a word" but "boundary between word and non-word"."a\\b", and see other comments about\b), your groups will always containaonly. So, what is it you want to do?