Replacing regex with the same amount of "." as its length

Question

See this for my current attempt: http://regexr.com?374vg

I have a regex that captures what I want it to capture, the thing is that the String().replaceAll("regex", ".") replaces everything with just one ., which is fine if it's at the end of the line, but otherwise it doesn't work.

How can I replace every character of the match with a dot, so I get the same amount of . symbols as its length?

Answer 1

Here's a one line solution:

str = str.replaceAll("(?<=COG-\\d{0,99})\\d", ".").replaceAll("COG-(?=\\.+)", "....");

Here's some test code:

String str = "foo bar COG-2134 baz";
str = str.replaceAll("(?<=COG-\\d{0,99})\\d", ".").replaceAll("COG-(?=\\.+)", "....");
System.out.println(str);

Output:

foo bar ........ baz

Answer 2

This is not possible using String#replaceAll. You might be able to use Pattern.compile(regexp) and iterate over the matches like so:

StringBuilder result = new StringBuilder();
Pattern pattern = Pattern.compile(regexp);
Matcher matcher = pattern.matcher(inputString);
int previous = 0;
while (matcher.find()) {
    result.append(inputString.substring(previous, matcher.start()));
    result.append(buildStringWithDots(matcher.end() - matcher.start()));
    previous = matcher.end();
}
result.append(inputString.substring(previous, inputString.length()));

To use this you have to define buildStringWithDots(int length) to build a String containing length dots.