c# - Regex that removes the 2 trailing letters from a string not preceded with other letters -

this in c#. i've been bugging head not luck far.

so example

123456bvc --> 123456bvc (keep same) 123456bv --> 123456 (remove trailing letters)  12345v -- > 12345v (keep same) 12345 --> 12345 (keep same) abc123ab --> abc123 (remove trailing letters)  

it can start anything.

i've tried @".*[a-za-z]{2}$" no luck

this in c# return string removing 2 trailing letters if exist , not preceded letter.

match result = regex.match(mystring, pattern); return result.value; 

your @".*[a-za-z]{2}$" regex matches 0+ characters other newline (as many possible) , 2 ascii letters @ end of string. not check context, 2 letters matched regardless of comes before them.

you need regex match last 2 letters not preceded letter:

(?<!\p{l})\p{l}{2}$ 

see this regex demo.

details:

• (?<!\p{l}) - fails match if letter (\p{l}) found before current position (you may use [a-za-z] if want deal ascii letters)
• \p{l}{2} - 2 letters
• $ - end of string.

in c#, use

var result = regex.replace(mystring, @"(?<!\p{l})\p{l}{2}$", string.empty);