Regex to get everything but last chars of capture group

Question

How do I use regex to select everything before last 4 char in a capture group?

Example:

String str = "{Index1=StudentData(studentName=Sam, idNumber=321231312), Index2=StudentData(studentName=Adam, idNumber=5675), Index3=StudentData(studentName=Lisa, idNumber=67124)}";
String regex = "(?<=idNumber=)[a-zA-Z1-9]+(?=\))";

System.out.println(str.replaceAll(regex, "*"));

Current output:

{Index1=StudentData(studentName=Sam, idNumber=*), Index2=StudentData(studentName=Adam, idNumber=*), Index3=StudentData(studentName=Lisa, idNumber=*)}

Desired output:

{Index1=StudentData(studentName=Sam, idNumber=*****1312), Index2=StudentData(studentName=Adam, idNumber=5675), Index3=StudentData(studentName=Lisa, idNumber=*7124)

Answer 1

You can use this regex in Java:

(\hidNumber=|(?!^)\G)[a-zA-Z1-9](?=[a-zA-Z1-9]{4,}\))

And replace with $1*.

RegEx Demo

Java Code:

final String re = "(\\hidNumber=|(?!^)\\G)[a-zA-Z1-9](?=[a-zA-Z1-9]{4,}\\));
String r = s.replaceAll(re, "$1*");

Breakdown:

• (: Start capture group #1
  • \h: Match a whitespace
  • idNumber=: Match text idNumber=
  • |: OR
  • (?!^)\G: Start at the end of the previous match
• ): Close capture group #1
• [a-zA-Z1-9]: Match an ASCII letter or digit 1-9
• (?=[a-zA-Z1-9]{4,}\)): Make sure that ahead of current position we have at least 4 ASCII letters or digits 1-9 followed by )