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Escaping, special characters

As we've seen, a backslash "\" is used to denote character classes. So it's a special character.

There are other special characters as well, that have special meaning in a regexp. They are used to do more powerful searches.

Here's a full list of them: pattern:[ \ ^ $ . | ? * + ( ).

Don't try to remember it -- when we deal with each of them separately, you'll know it by heart automatically.


To use a special character as a regular one, prepend it with a backslash.

That's also called "escaping a character".

For instance, we need to find a dot pattern:'.'. In a regular expression a dot means "any character except a newline", so if we really mean "a dot", let's put a backslash before it: pattern:\..

alert( "Chapter 5.1".match(/\d\.\d/) ); // 5.1

Parentheses are also special characters, so if we want them, we should use pattern:\(. The example below looks for a string "g()":

alert( "function g()".match(/g\(\)/) ); // "g()"

If we're looking for a backslash \, then we should double it:

alert( "1\\2".match(/\\/) ); // '\'

A slash

The slash symbol '/' is not a special character, but in JavaScript it is used to open and close the regexp: pattern:/...pattern.../, so we should escape it too.

Here's what a search for a slash '/' looks like:

alert( "/".match(/\//) ); // '/'

From the other hand, the alternative new RegExp syntaxes does not require escaping it:

alert( "/".match(new RegExp("/")) ); // '/'

new RegExp

If we are creating a regular expression with new RegExp, then we need to do some more escaping.

For instance, consider this:

let reg = new RegExp("\d\.\d");

alert( "Chapter 5.1".match(reg) ); // null

It doesn't work, but why?

The reason is string escaping rules. Look here:

alert("\d\.\d"); // d.d

Backslashes are used for escaping inside a string and string-specific special characters like \n. The quotes "consume" and interpret them, for instance:

  • \n -- becomes a newline character,
  • \u1234 -- becomes the Unicode character with such code,
  • ...And when there's no special meaning: like \d or \z, then the backslash is simply removed.

So the call to new RegExp gets a string without backslashes.

To fix it, we need to double backslashes, because quotes turn \\ into \:

let regStr = "\\d\\.\\d";
alert(regStr); // \d\.\d (correct now)

let reg = new RegExp(regStr);

alert( "Chapter 5.1".match(reg) ); // 5.1