Java Regex Russian Initials

Question

О.Г.дов.

I'm trying to match the "О.Г ." in this fragment. Normally, initials should be without spaces, so I need a regex that can account for optional spaces.

I've been using:

[А-Я]\s*\.\s*[А-Я]\s*\.\s*

But it seems to not match this correctly. I'm unsure why.Note, I am typing \\ in java before "s" and ".". Could someone spot an error?

Are you using `matches()`? Use `find()`, and add `^` at the start. — Wiktor Stribiżew, Feb 16 '18 at 22:59
That worked, I'll have to look into the difference.Thank you! — dj1121, Feb 16 '18 at 23:02
`[А-Я]` does not include 1 Russian letter, do not forget to include it there. — Wiktor Stribiżew, Feb 16 '18 at 23:17

Przemysław Moskal · Accepted Answer · 2018-02-16T23:13:05.730

1

Regex to try:

"([А-Я]\\s*\\.\\s*[А-Я]\\s*\\.)\\s*.*"

Explanation and sample code:

If Matcher finds a match for given regex, you get rid of all unnecessary spaces in captured group #1 (in the parenthesis ()) and print it to the console:

String source = "О . Г . дов.";
Pattern p = Pattern.compile("(^[А-Я]\\s*\\.\\s*[А-Я]\\s*\\.)\\s*.*");
Matcher m = p.matcher(source);
if(m.find()) {
    String resultWithoutSpaces = m.group(1).replaceAll(" ", "");
    System.out.println(resultWithoutSpaces);
}

Output you get:

О.Г.

edited Feb 16 '18 at 23:13

answered Feb 16 '18 at 23:03

Przemysław Moskal

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@dj1121 Glad that I could help - please have a look at my latest edit, I improved my answer to make it remove all unnecessary spaces from result initials. – Przemysław Moskal Feb 16 '18 at 23:14

Java Regex Russian Initials

1 Answers1

Regex to try:

Explanation and sample code: