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 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.
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: .
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: .
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.
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?]
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?]
In a regular expression that is defined 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.
.gif) Note: |
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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 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.