Your regular expression [^a-zA-Z0-9]\s/g says match any character that is not a number or letter followed by a space.

Remove the \s and you should get what you are after if you want a _ for every special character.

var newString = str.replace(/[^A-Z0-9]/ig, "_");

That will result in hello_world___hello_universe

If you want it to be single underscores use a + to match multiple

var newString = str.replace(/[^A-Z0-9]+/ig, "_");

That will result in hello_world_hello_universe

You should use the string replace function, with a single regex. Assuming by special characters, you mean anything that's not letter, here is a solution:

const str = "abc's test#s";
console.log(str.replace(/[^a-zA-Z ]/g, ""));

Some issues:

• The inner condition should in fact be the opposite: you want to do nothing when it is a punctuation character, i.e. you don't want to add it to the result. Only in the other case you want to do that.

• The call !compress(i) is wrong: first of all that function expects a string, not an index, and it returns a string, not a boolean (so to perform ! on it). It seems like you want to call your function recursively, and although that is an option, you are also iterating over the string. You should do one of the two: recursion or iteration.

• You reference a variable ch in the compress function which you have not defined there.

So, if you want to write compress the iteration way, change your code as follows:

var compress = function(s) {
    var result = "", ch; // define ch.

    //loop to traverse s
    for (var i = 0; i < s.length; i++) {
        ch = s[i]; // initialise ch.
        if (!isPunct(ch)) result = result + ch; // only add when not punctuation
    }
    return result;
}

If on the other hand you want to keep your recursive call to compress, then you should do away with your for loop:

var compress = function(s) {
    var result = "", ch, rest;

    if (s.length == 0) return '';
    result = compress(s.substr(1)); // recursive call
    ch = s[0];
    if (!isPunct(ch)) result = ch + result;
    return result;
}

The function isPunct also has a strange thing happening: it assigns a boolean value to ch in the if expression. This does not make your function malfunction, but that assignment serves no purpose: the match method already returns the boolean you need for your if condition.

It is also not really nice-looking to first evaluate a boolean expression in an if to then return that same value in the form of false and true. This you can do by just returning the evaluated expression itself:

var isPunct = function(ch) {
    return ch.match(/[,.!?;:'-]/g);
}

On a final note, you don't really need the isPunct function if you only use it in compress. The whole logic can be performed in one function only, like this:

let compress = s => s.replace(/[,.!?;:'-]/g,'');
// Demo:  
console.log(compress('a,b,c')); // abc

If you prefer to keep isPunct and don't want to repeat the regular expression elsewhere, then you can do the replace like this:

let isPunct = ch => ch.match(/[,.!?;:'-]/g);

let compress = s => Array.from(s).filter(ch => !isPunct(ch)).join('');
// Demo:  
console.log(compress('a,b,c')); // abc

Note how the use of ES6 arrow functions and ES5 Array methods makes the code quite lean.

To replace multiple spaces with a single space , you can try below regex :

/  +/g

Here, the Regex detects 1 or more space(s) throughout the string.

We need to use the regex in the replace method as shown below:

let myValue= 'Premium+-+Mobile+Plan' 
console.log(myValue); // Premium+-+Mobile+Plan 

let newValue = myValue.replace(/[+-]/g, ' ').replace(/  +/g, ' ');
console.log(newValue ); // Premium Mobile Plan

You can do it as:

var str = "Multiple Choice";
var strLower = str.toLowerCase();

strLower.replace(/\s/g, '');

Working demo.

The Regex

\s is the regex for "whitespace", and g is the "global" flag, meaning match all \s (whitespaces).

Try with this function:

function removeSpecialChars(str) {
  return str.replace(/(?!\w|\s)./g, '')
    .replace(/\s+/g, ' ')
    .replace(/^(\s*)([\W\w]*)(\b\s*$)/g, '$2');
}

• 1st regex /(?!\w|\s)./g remove any character that is not a word or whitespace. \w is equivalent to [A-Za-z0-9_]
• 2nd regex /\s+/g find any appearance of 1 or more whitespaces and replace it with one single white space
• 3rd regex /^(\s*)([\W\w]*)(\b\s*$)/g trim the string to remove any whitespace at the beginning or the end.

You should just be able to use not in the matching group and remove everything that isn't a letter:

/[^a-zA-Z]/g

DEMO

replace(/[^a-z0-9\s]/gi, '') will filter the string down to just alphanumeric values and replace(/[_\s]/g, '-') will replace underscores and spaces with hyphens:

str.replace(/[^a-z0-9\s]/gi, '').replace(/[_\s]/g, '-')

Source for Regex: RegEx for Javascript to allow only alphanumeric

Here is a demo: http://jsfiddle.net/vNfrk/