You can use regex

myString.replace(/[^\w\s!?]/g,'');

This will replace everything but a word character, space, exclamation mark, or question.

Character Class: \w stands for "word character", usually [A-Za-z0-9_]. Notice the inclusion of the underscore and digits.

\s stands for "whitespace character". It includes [ \t\r\n].

If you don't want the underscore, you can use just [A-Za-z0-9].

myString.replace(/[^A-Za-z0-9\s!?]/g,'');

For unicode characters, you can add something like \u0000-\u0080 to the expression. That will exclude all characters within that unicode range. You'll have to specify the range for the characters you don't want removed. You can see all the codes on Unicode Map. Just add in the characters you want kept or a range of characters.

For example:

myString.replace(/[^A-Za-z0-9\s!?\u0000-\u0080\u0082]/g,'');

This will allow all the previously mentioned characters, the range from \u0000-\u0080 and \u0082. It will remove \u0081.

A carefully chosen regular expression can strip out the weightings in one replacement. A second switches the hyphens - for underscores _

const locales = 'fr-CH, fr;q=0.9, en;q=0.8, de;q=0.7, *;q=0.5';
newLocales = locales.replace(/;q=\d*\.\d*/g,'').replace(/-/g,'_');
console.log(newLocales);   //  fr_CH, fr, en, de, *

To remove characters, you'd need to use something that actually does that, like the string replace function (which can accept a regular expression as the "from" parameter).

Then you're just dealing with a normal application of a character class, which in JavaScript (and most other regular expression variants) is described using [...], where ... is what should be in the class. You'd use the ^ at the beginning to invert the meaning of the class:

In your case, it might be:

str = str.replace(/[^A-Za-z']/g, "");

...which will replace except the English characters A-Z (ABCDEFGHIJKLMNOPQRSTUVWXYZ), a-z (abcdefghijklmnopqrstuvwxyz), and the single quote with nothing (e.g., remove it).

let str = "This is a test with the numbers 123 and a '.";

console.log("before:", str);
str = str.replace(/[^A-Za-z']/g, "");
console.log("after: ", str);

However, note that alphabetic characters not used in English will not be excepted, and there are a lot of those in the various languages used on the web (and even, perversely, in English, in "borrowed" words like "voilà" and "naïve").

You've said you're okay with just English A-Z, but for others coming to this: In environemnts supporting ES2018 and above's Unicode property matching, you could handle anything considered "alphabetic" by Unicode instead of just A-Z by using the \p{Alpha} property. The \p means "matching this Unicode property" (as usual, the lowercase version \p means "matching" and the uppercase version \P means "not matching") and the {Alpha} means "alphabetic":

str = str.replace(/[^\p{Alpha}']/gu, "");

(Note that, again, \p{Alpha} means "alphabetic" but because it's in a negated character class, we're excluding alphabetic characters.)

Note the u flag on that, to enable newer Unicode features. That handles the "voilà" and "naïve" examples too:

let str = "This is a test with the numbers 123 and a ' and voilà and naïve.";

console.log("before:", str);
str = str.replace(/[^\p{Alpha}']/gu, "");
console.log("after: ", str);

You can use \D which means non digits.

var removedText = self.val().replace(/\D+/g, '');

jsFiddle.

You could also use the HTML5 number input.

<input type="number" name="digit" />

jsFiddle.

Add the global option at the end of the regex - see demo below:

console.log("AkjkljKK".replace(/(?![A-Z])./g, ''));

The :space: portion of the regex makes no sense, and probably does not do what you intend.

> x <- "abc:def."
> gsub("[^a-zA-Z0-9,-:space:]", " ", x, perl = TRUE)
[1] "abc:def."

Notice that the colon and period are still present after the substitution.

In fact, inside the character class, ,-: means "all characters with ASCII codes from 44 (the comma) up to 58 (the colon)".

A literal hyphen must be the first or the last character in a character class; otherwise, it is treated as a range (like A-Z).

If you want a character class for whitespace, use "\\s" or [:space:].

So, if you wanted to convert all consecutive strings of junk to a single space, preserving only letters, digits, commas, slashes, hyphens, and whitespace, you could write:

gsub("[^-,/a-zA-Z0-9[:space:]]+", " ", x, perl = TRUE)

or

gsub("[^-,/a-zA-Z0-9\\s]+", " ", x, perl = TRUE)

You can write a negated character class with ^ to include the period, ., and the set of digits, 0-9:

var z = x.replace(/[^0-9.]/g, '');

z is then "100.00".

Use negated character set:

If you're okay keeping the underscore, you can use:

str.replace(/[^\w]/g, "-") // \w is the same as [A-Za-z0-9_]

If you want to replace that as well, then you can use:

str.replace(/[^A-Za-z0-9]/g, "-")

Edit: As mentioned by @JamesBuck, there are actually negated shorthands: [^\w] is the same as \W, so if you don't need to replace the underscore, you can use that, resulting in:

str.replace(/\W/g, "-")

here is a regular expression that will match every thing that is not a number or a space ..

[^0-9\s]

explanation hey regex engine :

1- match every thing .

2- ^ --> that's not

3- 0-9 --> a number

4- \s --> and a space

There's probably a more concise regex, but this will certainly work:

string.replaceAll("[^a-zA-Z0-9]", "");

Removing non-alphanumeric chars

The following is the/a correct regex to strip non-alphanumeric chars from an input string:

input.replace(/\W/g, '')

Note that \W is the equivalent of [^0-9a-zA-Z_] - it includes the underscore character. To also remove underscores use e.g.:

input.replace(/[^0-9a-z]/gi, '')

The input is malformed

Since the test string contains various escaped chars, which are not alphanumeric, it will remove them.

A backslash in the string needs escaping if it's to be taken literally:

"\\test\\red\\bob\\fred\\new".replace(/\W/g, '')
"testredbobfrednew" // output

Handling malformed strings

If you're not able to escape the input string correctly (why not?), or it's coming from some kind of untrusted/misconfigured source - you can do something like this:

JSON.stringify("\\test\red\bob\fred\new").replace(/\W/g, '')
"testredbobfrednew" // output

Note that the json representation of a string includes the quotes:

JSON.stringify("\\test\red\bob\fred\new")
""\\test\red\bob\fred\new""

But they are also removed by the replacement regex.

As @chris85 said, you can use the regular expression [^0-9a-z-A-Z] to replace all the characters that are not letters, numbers, or whitespace.

Here is a function that will do what you want:

function clean(str) {
    return str.replace(/[^0-9a-z-A-Z ]/g, "").replace(/ +/, " ")
}

The second replace call is needed to remove the runs of extra whitespace that are made from removing a character that is between spaces.