Question:
I need to insert into a regular expression pattern:
$@"^(?:[^\p{{L}}]|[{exclusion}])+$" //Цель: Запретить использование каких либо букв в строке, кроме тех что заданы в переменной - exclusion
string variable:
string exclusion;
in which all control characters would be escaped, which would avoid errors associated with the operation of the regular expression.
I found the Regex.Escape() method. But it doesn't suit my needs. For example, if exclusion = @"[text]" passed to the Regex.Escape() method, it will return the string "\\[text]" . After inserting this line into the pattern, instead of the exclusion variable:
$@"^(?:[^\p{{L}}]|[{exclusion}])+$" //Цель: запретить использование каких либо букв в строке, кроме тех что заданы в переменной - exclusion
it takes the following form:
$@"^(?:[^\p{{L}}]|[\[text]])+$"
As a result, the regular expression does not work correctly. I suspect that the reason is the extra character – ]
Can you please tell me how to escape all control characters in a string? Is there any other way besides the Regex.Escape() method? Maybe I somehow used it incorrectly and do not notice my mistake?
Answer:
Regex.Escape escapes those characters that are considered special outside of character classes:
Escapes a minimal set of characters (
\,*,+,?,|,{,[,(,),^,$,.,#, and white space) by replacing them with their escape codes. ( Escapes the minimum set of characters (\,*,+,?,|,{,[,(,),^,$,.,#and whitespace) by replacing them with their escape codes )
In fact, only the following characters are considered special within character classes:
-
^– can mean an exclusive type of a character class if it is immediately after the opening[ -
]– closes the character class -
\– escapes special characters -
-– specifies a range of characters or "character class subtraction"
To escape these characters, it is enough to use
exclusion.Replace("\\", @"\\").Replace("^", @"\^").Replace("-", @"\-").Replace("]", @"\]")
or
Regex.Replace(exclusion, @"[]^\\-]", "\\$&")
Solution:
var pattern = $@"^(?:[^\p{{L}}]|[{Regex.Replace(exclusion, @"[]^\\-]", "\\$&")}])+$";
Or (since [^\p{L}] = \P{L} ):
var pattern2 = $@"^[\P{{L}}{Regex.Replace(exclusion, @"[]^\\-]", "\\$&")}]+$";