Regex.Escape(String) Method
Definition
Important
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Escapes a minimal set of characters (\, *, +, ?, |, {, [, (,), ^, $, ., #, and white space) by replacing them with their escape codes. This instructs the regular expression engine to interpret these characters literally rather than as metacharacters.
public:
static System::String ^ Escape(System::String ^ str);public static string Escape (string str);static member Escape : string -> stringPublic Shared Function Escape (str As String) As StringParameters
- str
- String
The input string that contains the text to convert.
Returns
A string of characters with metacharacters converted to their escaped form.
Exceptions
str is null.
Examples
The following example extracts comments from text. It assumes that the comments are delimited by a begin comment symbol and an end comment symbol that is selected by the user. Because the comment symbols are to be interpreted literally, they are passed to the Escape method to ensure that they cannot be misinterpreted as metacharacters. In addition, the example explicitly checks whether the end comment symbol entered by the user is a closing bracket (]) or brace (}). If it is, a backslash character (\) is prepended to the bracket or brace so that it is interpreted literally. Note that the example also uses the Match.Groups collection to display the comment only, rather than the comment together with its opening and closing comment symbols.
using System;
using System.Text.RegularExpressions;
public class Example
{
public static void Main()
{
ConsoleKeyInfo keyEntered;
char beginComment, endComment;
Console.Write("Enter begin comment symbol: ");
keyEntered = Console.ReadKey();
beginComment = keyEntered.KeyChar;
Console.WriteLine();
Console.Write("Enter end comment symbol: ");
keyEntered = Console.ReadKey();
endComment = keyEntered.KeyChar;
Console.WriteLine();
string input = "Text [comment comment comment] more text [comment]";
string pattern;
pattern = Regex.Escape(beginComment.ToString()) + @"(.*?)";
string endPattern = Regex.Escape(endComment.ToString());
if (endComment == ']' || endComment == '}') endPattern = @"\" + endPattern;
pattern += endPattern;
MatchCollection matches = Regex.Matches(input, pattern);
Console.WriteLine(pattern);
int commentNumber = 0;
foreach (Match match in matches)
Console.WriteLine("{0}: {1}", ++commentNumber, match.Groups[1].Value);
}
}
// The example shows possible output from the example:
// Enter begin comment symbol: [
// Enter end comment symbol: ]
// \[(.*?)\]
// 1: comment comment comment
// 2: comment
Imports System.Text.RegularExpressions
Module Example
Public Sub Main()
Dim keyEntered As ConsoleKeyInfo
Dim beginComment, endComment As Char
Console.Write("Enter begin comment symbol: ")
keyEntered = Console.ReadKey()
beginComment = keyEntered.KeyChar
Console.WriteLine()
Console.Write("Enter end comment symbol: ")
keyEntered = Console.ReadKey()
endComment = keyEntered.KeyChar
Console.WriteLine()
Dim input As String = "Text [comment comment comment] more text [comment]"
Dim pattern As String = Regex.Escape(beginComment.ToString()) + "(.*?)"
Dim endPattern As String = Regex.Escape(endComment.ToString())
If endComment = "]"c OrElse endComment = "}"c Then endPattern = "\" + endPattern
pattern += endPattern
Dim matches As MatchCollection = Regex.Matches(input, pattern)
Console.WriteLine(pattern)
Dim commentNumber As Integer = 0
For Each match As Match In matches
commentNumber += 1
Console.WriteLine("{0}: {1}", commentNumber, match.Groups(1).Value)
Next
End Sub
End Module
' The example shows possible output from the example:
' Enter begin comment symbol: [
' Enter end comment symbol: ]
' \[(.*?)\]
' 1: comment comment comment
' 2: comment
Remarks
Escape converts a string so that the regular expression engine will interpret any metacharacters that it may contain as character literals. For example, consider a regular expression that is designed to extract comments that are delimited by straight opening and closing brackets ([ and ]) from text. In the following example, the regular expression "[(.*?)]" is interpreted as a character class. Rather than matching comments embedded in the input text, the regular expression matches each opening or closing parenthesis, period, asterisk, or question mark.
string pattern = "[(.*?)]";
string input = "The animal [what kind?] was visible [by whom?] from the window.";
MatchCollection matches = Regex.Matches(input, pattern);
int commentNumber = 0;
Console.WriteLine("{0} produces the following matches:", pattern);
foreach (Match match in matches)
Console.WriteLine(" {0}: {1}", ++commentNumber, match.Value);
// This example displays the following output:
// [(.*?)] produces the following matches:
// 1: ?
// 2: ?
// 3: .
Dim pattern As String = "[(.*?)]"
Dim input As String = "The animal [what kind?] was visible [by whom?] from the window."
Dim matches As MatchCollection = Regex.Matches(input, pattern)
Dim commentNumber As Integer = 0
Console.WriteLine("{0} produces the following matches:", pattern)
For Each match As Match In matches
commentNumber += 1
Console.WriteLine("{0}: {1}", commentNumber, match.Value)
Next
' This example displays the following output:
' 1: ?
' 2: ?
' 3: .
However, if the opening bracket is escaped by passing it to the Escape method, the regular expression succeeds in matching comments that are embedded in the input string. The following example illustrates this.
string pattern = Regex.Escape("[") + "(.*?)]";
string input = "The animal [what kind?] was visible [by whom?] from the window.";
MatchCollection matches = Regex.Matches(input, pattern);
int commentNumber = 0;
Console.WriteLine("{0} produces the following matches:", pattern);
foreach (Match match in matches)
Console.WriteLine(" {0}: {1}", ++commentNumber, match.Value);
// This example displays the following output:
// \[(.*?)] produces the following matches:
// 1: [what kind?]
// 2: [by whom?]
Dim pattern As String = Regex.Escape("[") + "(.*?)]"
Dim input As String = "The animal [what kind?] was visible [by whom?] from the window."
Dim matches As MatchCollection = Regex.Matches(input, pattern)
Dim commentNumber As Integer = 0
Console.WriteLine("{0} produces the following matches:", pattern)
For Each match As Match In matches
commentNumber += 1
Console.WriteLine(" {0}: {1}", commentNumber, match.Value)
Next
' This example displays the following output:
' \[(.*?)] produces the following matches:
' 1: [what kind?]
' 2: [by whom?]
In a regular expression that is defined by using static text, characters that are to be interpreted literally rather than as metacharacters can be escaped by preceding them with a backslash symbol (\) as well as by calling the Escape method. In a regular expression that is defined dynamically using characters that are not known at design time, calling the Escape method is particularly important to ensure that the regular expression engine interprets individual characters as literals rather than as metacharacters.
Note
If a regular expression pattern includes either the number sign (#) or literal white-space characters, they must be escaped if input text is parsed with the RegexOptions.IgnorePatternWhitespace option enabled.
While the Escape method escapes the straight opening bracket ([) and opening brace ({) characters, it does not escape their corresponding closing characters (] and }). In most cases, escaping these is not necessary. If a closing bracket or brace is not preceded by its corresponding opening character, the regular expression engine interprets it literally. If an opening bracket or brace is interpreted as a metacharacter, the regular expression engine interprets the first corresponding closing character as a metacharacter. If this is not the desired behavior, the closing bracket or brace should be escaped by explicitly prepending the backslash (\) character. For an illustration, see the Example section.
