Replace digits of length 2 to 8 with a specific character using regex
We have a security issue where a certain field in the database contains some sensitive information. I need a way to find numbers from 2 to 8, replace the digits with "filler" of the same length.
For example:
Jim8888Dandy
Mike9999999999Thompson * Note: this is 10 in length and we don't want to replace the digits
123Area Code
Tim Johnson5555555
In these cases, at any time we find a number that is between 2 and 8 (inclusive), then I want to replace / fill / replace this value with 0 and keep the length of the original digits
Final result
Jim0000Dandy
Mike9999999999Thompson
000Area Code
Tim Johnson0000000
Is there an easy way to accomplish this with RegEx?
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You need to provide a static evaluator method to replace. It replaces the numbers in the match with zeros:
public static string Evaluate(Match m)
{
return Regex.Replace(m.Value, "[0-9]", "0");
}
And then use it with this code:
string input = "9999999099999Thompson534543";
MatchEvaluator evaluator = new MatchEvaluator(Program.Evaluate);
string replaced = Regex.Replace(input, "(?:^|[^0-9])[0-9]{2,8}(?:$|[^0-9])", evaluator);
Regular expression:
-
(?:^|[^0-9])
- must start or start with a non-digit -
[0-9]{2,8}
- capture 2 to 8 digits -
(?:$|[^0-9])
- must be at the end or after it not contain numbers
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For the clever regex department only. This is not an efficient regular expression.
(?<=(?>(?'front'\d){0,7}))\d(?=(?'back'(?'-front'\d)){0,7}(?!\d))((?'-front')|(?'-back'))
Replace with 0
.
/(?<=(?>(?'front'\d){0,7})) # Measure how many digits we're behind.
\d # This digit is matched
(?=
(?'back' # Measure how many digits we're in front of.
(?'-front'\d)){0,7}
# For every digit here, subtract one group from 'front',
# As to assert we'll never go over the < 8 digit requirement.
(?!\d) # no more digits
)
(
(?'-front') # At least one capturing group left for 'front' or 'back'
|(?'-back') # for > 2 digits requirement.
)/x
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