var string = "    This    should  become   something          else   too . ";
string = string.replace(/\s+/g, " ");

This code replaces a consecutive set of whitespace characters (\s+) by a single white space. Note that a white-space character also includes tab and newlines. Replace \s by a space if you only want to replace spaces.

If you also want to remove the whitespace at the beginning and end, include:

string = string.replace(/^\s+|\s+$/g, "");

This line removes all white-space characters at the beginning (^) and end ($). The g at the end of the RegExp means: global, ie match and replace all occurences.

Given that you also want to cover tabs, newlines, etc, just replace \s\s+ with ' ':

string = string.replace(/\s\s+/g, ' ');

If you really want to cover only spaces (and thus not tabs, newlines, etc), do so:

string = string.replace(/  +/g, ' ');

I would recommend you use the literal notation, and the \s character class:

//..
return str.replace(/\s/g, '');
//..

There's a difference between using the character class \s and just ' ', this will match a lot more white-space characters, for example '\t\r\n' etc.., looking for ' ' will replace only the ASCII 32 blank space.

The RegExp constructor is useful when you want to build a dynamic pattern, in this case you don't need it.

Moreover, as you said, "[\s]+" didn't work with the RegExp constructor, that's because you are passing a string, and you should "double escape" the back-slashes, otherwise they will be interpreted as character escapes inside the string (e.g.: "\s" === "s" (unknown escape)).

When the regex matches the first time (on "A B"), this part of the string in consumed by the engine, so it is not matched again, even though your regex has the global ('g') flag.

You could achieve the expected result by using a positive lookahead ((?=PATTERN)) instead, that won't consume the match:

value = "Hello I B M"
value = value.replace(/([A-Z])\s(?=[A-Z])/g, '$1')
console.log(value) // Prints "Hello IBM"

To make it not remove the space if the next uppercase letter is the first in a word, you can increment the lookahead pattern with using a word boundary \b to make that restriction:

value = "Hello I B M Dude"
value = value.replace(/([A-Z])\s(?=[A-Z]\b)/g, '$1')
console.log(value) // Prints "Hello IBM Dude"

Note: As @CasimirHyppolite noted, the following letter has to be made optional, or the second regex won't work if the last character of the string is uppercase. Thus, the pattern ([^A-Za-z]|$), which can be read as "not a letter, or the end of the string".

Edit: Simplify lookahead from (?=A-Z) to (?=[A-Z]\b) as suggested by @hwnd

You can use the regex /\s{2,}/g:

var s = "Be an      excellent     person"
s.replace(/\s{2,}/g, ' ');