SyntaxError: \ at end of pattern

The JavaScript exception "\ at end of pattern" occurs when a regular expression pattern ends with an unescaped backslash (\). In a regex literal, the backslash would cause the closing slash / to be a literal character, so this can only appear when using the {{jsxref("RegExp/RegExp", "RegExp()")}} constructor.

Message#

SyntaxError: Invalid regular expression: /\/: \ at end of pattern (V8-based)
SyntaxError: \ at end of pattern (Firefox)
SyntaxError: Invalid regular expression: \ at end of pattern (Safari)

Error type#

{{jsxref("SyntaxError")}}

What went wrong?#

A backslash cannot appear literally in a regular expression. It either precedes another character to escape it, or is itself escaped by another backslash. A backslash at the end of a regular expression pattern is invalid because it is not escaping anything.

Examples#

Double-escaping backslashes#

This error can only happen when using the RegExp() constructor. Consider the following code, which intends to match a single backslash character:

```js example-bad const pattern = new RegExp("\");

In JavaScript strings, backslashes are also escape sequences. Therefore, the two backslashes in the string literal `"\\"` are interpreted as a single backslash. The `RegExp()` constructor then only sees a single backslash character in the regex source. To fix this, you need to double-escape the backslash:

```js example-good
const pattern = new RegExp("\\\\");

The four backslashes in the string literal represent two backslashes in the regex source, which then becomes a character escape for a single backslash literal character.

See also#

• Regular expressions
• Escape sequences