I am trying to write a regular expression to mask an email address. Example below.

input: john.doe@example.en.com

output: j*******@e*********.com

I have tried the following but I just can't seem to get it working correctly.

regex:(?<=.).(?=[^@]\*?@)

output:j*******@example.en.com

regex:(?<=.).(?=[^@]\*?)(?=[^\.]\*?\.)

output:j******************.com

Any help would be appreciated. demo

Update with various masking email solutions

foo@bar.com ⇒ f**@b**.com (current question) - s.replaceAll("(?<=.)[^@](?=[^@]*?@)|(?:(?<=@.)|(?!^)\\G(?=[^@]*$)).(?=.*\\.)", "*") (see the regex demo)
foo@bar.com ⇒ f**@b*r.com - s.replaceAll("(?<=.)[^@](?=[^@]*?@)|(?:(?<=@.)|(?!^)\\G(?=[^@]*$)).(?=.*[^@]\\.)", "*") (see the regex demo)
foo@bar.com ⇒ f*o@b*r.com - s.replaceAll("(?<=.)[^@](?=[^@]*?[^@]@)|(?:(?<=@.)|(?!^)\\G(?=[^@]*$)).(?=.*[^@]\\.)", "*") (see the regex demo)
foo@bar.com ⇒ f**@b*****m - s.replaceAll("(?<=.)[^@](?=[^@]*?@)|(?:(?<=@.)|(?!^)\\G(?=[^@]*$)).(?!$)", "*") (see the regex demo)
foo@bar.com ⇒ f*o@b*****m - s.replaceAll("(?<=.)[^@](?=[^@]*[^@]@)|(?:(?<=@.)|(?!^)\\G(?=[^@]*$)).(?!$)", "*") (see the regex demo)

Original answer

In case you can't use a code-based solution, you may use

s.replaceAll("(?<=.)[^@](?=[^@]*?@)|(?:(?<=@.)|(?!^)\\G(?=[^@]*$)).(?=.*\\.)", "*")

What it does:

(?<=.)[^@](?=[^@]*?@) -any char other than @ ([^@]) that is preceded by any single char ((?<=.)) and is followed with any 0 or more chars other than @ up to a @ ((?=[^@]*?@))
| - or
(?:(?<=@.)|(?!^)\\G(?=[^@]*$)) - match a location in the string that is preceded with @ and any char ((?<=@.)) or (|) the end of the previous successful match ((?!^)\\G) that is followed with any 0+ chars other than @ uo to the end of string ((?=[^@]*$))
. - any single char
(?=.*\\.) - followed with any 0+ chars up to the last . symbol in the string.