3.4 KiB
Escaping, special characters
As we've seen, a backslash "\" is used to denote character classes. So it's a special character in regexps (just like in a regular string).
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 the list -- soon we'll deal with each of them separately and you'll know them by heart automatically.
Escaping
Let's say we want to find a dot literally. Not "any character", but just a dot.
To use a special character as a regular one, prepend it with a backslash: pattern:\..
That's also called "escaping a character".
For example:
alert( "Chapter 5.1".match(/\d\.\d/) ); // 5.1 (match!)
alert( "Chapter 511".match(/\d\.\d/) ); // null (looking for a real dot \.)
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 \, it's a special character in both regular strings and regexps, so we should double it.
alert( "1\\2".match(/\\/) ); // '\'
A slash
A 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(/\//) ); // '/'
On the other hand, if we're not using /.../, but create a regexp using new RegExp, then we don't need to escape it:
alert( "/".match(new RegExp("/")) ); // '/'
new RegExp
If we are creating a regular expression with new RegExp, then we don't have to escape /, but need to do some other escaping.
For instance, consider this:
let reg = new RegExp("\d\.\d");
alert( "Chapter 5.1".match(reg) ); // null
It worked with pattern:/\d\.\d/, but with new RegExp("\d\.\d") it doesn't, why?
The reason is that backslashes are "consumed" by a string. Remember, regular strings have their own special characters like \n, and a backslash is used for escaping.
Please, take a look, what "\d.\d" really is:
alert("\d\.\d"); // d.d
The quotes "consume" backslashes 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. That's why it doesn't work!
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
Summary
- To search special characters
pattern:[ \ ^ $ . | ? * + ( )literally, we need to prepend them with\("escape them"). - We also need to escape
/if we're insidepattern:/.../(but not insidenew RegExp). - When passing a string
new RegExp, we need to double backslashes\\, cause strings consume one of them.