Question

I have the below groovy code

import java.util.regex.Matcher;
import java.util.regex.Pattern;

String myInput="License_All (12313)"
String myRegex="\\(\\d+\\)"


String ResultString
Pattern regex
Matcher regexMatcher

regex = Pattern.compile(myRegex, Pattern.DOTALL);

regexMatcher = regex.matcher(myInput);

List<String> matchList = new ArrayList<String>();
while (regexMatcher.find()) {
matchList.add(regexMatcher.group());
}

for(int i=0;i<matchList.size();i++)
{
myInput=myInput.replaceAll( matchList[i], '')
println matchList[i]
}
println myInput

It is supposed to remove (12313) part but instead it gives me the below output

(12313)
License_All ()

How do i remove all the bracket portion (12313)?

Answer 1

The matchList will contain the matched part, i.e. (12313). This one is then again taken as regular expression within replaceAll and therefore interpreted as group. Thus the brackets are not removed.

Instead you can use your regex directly in replaceAll:

myInput.replaceAll(myRegex, "")