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Showing content from https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp/escape below:

RegExp.escape() - JavaScript | MDN

Limited availability

The RegExp.escape() static method escapes any potential regex syntax characters in a string, and returns a new string that can be safely used as a literal pattern for the RegExp() constructor.

When dynamically creating a RegExp with user-provided content, consider using this function to sanitize the input (unless the input is actually intended to contain regex syntax). In addition, don't try to re-implement its functionality by, for example, using String.prototype.replaceAll() to insert a \ before all syntax characters. RegExp.escape() is designed to use escape sequences that work in many more edge cases/contexts than hand-crafted code is likely to achieve.

string

The string to escape.

A new string that can be safely used as a literal pattern for the RegExp() constructor. Namely, the following things in the input string are replaced:

• The first character of the string, if it's either a decimal digit (0–9) or ASCII letter (a–z, A–Z), is escaped using the \x character escape syntax. For example, RegExp.escape("foo") returns "\\x66oo" (here and after, the two backslashes in a string literal denote a single backslash character). This step ensures that if this escaped string is embedded into a bigger pattern where it's immediately preceded by \1, \x0, \u000, etc., the leading character doesn't get interpreted as part of the escape sequence.
• Regex syntax characters, including ^, $, \, ., *, +, ?, (, ), [, ], {, }, and |, as well as the / delimiter, are escaped by inserting a \ character before them. For example, RegExp.escape("foo.bar") returns "\\x66oo\\.bar", and RegExp.escape("(foo)") returns "\\(foo\\)".
• Other punctuators, including ,, -, =, <, >, #, &, !, %, :, ;, @, ~, ', `, and ", are escaped using the \x syntax. For example, RegExp.escape("foo-bar") returns "\\x66oo\\x2dbar". These characters cannot be escaped by prefixing with \ because, for example, /foo\-bar/u is a syntax error.
• The characters with their own character escape sequences: \f (U+000C FORM FEED), \n (U+000A LINE FEED), \r (U+000D CARRIAGE RETURN), \t (U+0009 CHARACTER TABULATION), and \v (U+000B LINE TABULATION), are replaced with their escape sequences. For example, RegExp.escape("foo\nbar") returns "\\x66oo\\nbar".
• The space character is escaped as "\\x20".
• Other non-ASCII line break and white space characters are replaced with one or two \uXXXX escape sequences representing their UTF-16 code units. For example, RegExp.escape("foo\u2028bar") returns "\\x66oo\\u2028bar".
• Lone surrogates are replaced with their \uXXXX escape sequences. For example, RegExp.escape("foo\uD800bar") returns "\\x66oo\\ud800bar".

TypeError

Thrown if string is not a string.

The following examples demonstrate various inputs and outputs for the RegExp.escape() method.

RegExp.escape("Buy it. use it. break it. fix it.");
// "\\x42uy\\x20it\\.\\x20use\\x20it\\.\\x20break\\x20it\\.\\x20fix\\x20it\\."
RegExp.escape("foo.bar"); // "\\x66oo\\.bar"
RegExp.escape("foo-bar"); // "\\x66oo\\x2dbar"
RegExp.escape("foo\nbar"); // "\\x66oo\\nbar"
RegExp.escape("foo\uD800bar"); // "\\x66oo\\ud800bar"
RegExp.escape("foo\u2028bar"); // "\\x66oo\\u2028bar"

The primary use case of RegExp.escape() is when you want to embed a string into a bigger regex pattern, and you want to ensure that the string is treated as a literal pattern, not as a regex syntax. Consider the following naïve example that replaces URLs:

function removeDomain(text, domain) {
  return text.replace(new RegExp(`https?://${domain}(?=/)`, "g"), "");
}

const input =
  "Consider using [RegExp.escape()](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp/escape) to escape special characters in a string.";
const domain = "developer.mozilla.org";
console.log(removeDomain(input, domain));
// Consider using [RegExp.escape()](/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp/escape) to escape special characters in a string.

Inserting the domain above results in the regular expression literal https?://developer.mozilla.org(?=/), where the "." character is a regex wildcard character. This means the string will match the string with any character in place of the ".", such as developer-mozilla-org. Therefore, it would incorrectly also change the following text:

const input =
  "This is not an MDN link: https://developer-mozilla.org/, be careful!";
const domain = "developer.mozilla.org";
console.log(removeDomain(input, domain));
// This is not an MDN link: /, be careful!

To fix this, we can use RegExp.escape() to ensure that any user input is treated as a literal pattern:

function removeDomain(text, domain) {
  return text.replace(
    new RegExp(`https?://${RegExp.escape(domain)}(?=/)`, "g"),
    "",
  );
}

Now this function will do exactly what we intend to, and will not transform developer-mozilla.org URLs.

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