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
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(?=[^@]*$)).(?=.*[^@]\\.)", "*") (see the regex demo)
foo@bar.com ⇒ f**@b*****m - s.replaceAll("(?<=.)[^@](?=[^@]*?@)|(?:(?<=@.)|(?!^)\\G(?=[^@]*$)).(?!$)", "*") (see the regex demo)
foo@bar.com ⇒ f*o@b*****m - s.replaceAll("(?<=.)[^@](?=[^@]*[^@]@)|(?:(?<=@.)|(?!^)\\G(?=[^@]*$)).(?!$)", "*") (see the regex demo)
In case you can't use a code-based solution, you may use
s.replaceAll("(?<=.)[^@](?=[^@]*?@)|(?:(?<=@.)|(?!^)\\G(?=[^@]*$)).(?=.*\\.)", "*")
See the regex demo
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.How about this one if you do not need the masks having the same number of characters of the original strings (which is more anonymous):
(?<=^.)[^@]*|(?<=@.).*(?=\.[^.]+$)
For example, if you replace the matches with ***, the result would be:
j***@e***.com