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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.
Escaping
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
\dor\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