Replace all occurrences of group

Question

I want to replace all the occurrences of a group in a string.

String test = "###,##.##0.0########";
System.out.println(test);
test = test.replaceAll("\\.0(#)", "0");
System.out.println(test);

The result I am trying to obtain is ###,##.##0.000000000 Basically, I want to replace all # symbols that are trailing the .0. I've found this about dynamic replacement but I can't really make it work.

The optimal solution will not take into account the number of hashes to be replaced (if that clears any confusion).

Answer 1

#(?!.*\\.0)

You can try this.Replace by 0.See demo.

https://regex101.com/r/yW3oJ9/12

Answer 2

You can use a simple regex to achieve your task.

#(?=#*+$)

(?=#*+$) = A positive look-ahead that checks for any # that is preceded by 0 or more # symbols before the end of string $. Edit: I am now using a possessive quantifier *+ to avoid any performance issues.

See demo

IDEONE:

String test = "###,##.##0.0###########################################";
test = test.replaceAll("#(?=#*+$)", "0");
System.out.println(test);

Answer 3

You can split your text on "0.0" and replace just for the second part:

String[] splited = "###,##.##0.0########".split("0.0");
String finalString = splited[0] + "0.0" + splited[1].replaceAll("#","0");