It was not asked precisely to remove accent (only special characters), but I needed to.

The solutions givens here works but they don’t remove accent: é, è, etc.

So, before doing epascarello’s solution, you can also do:

var newString = "développeur & intégrateur";

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

/**
 * Replaces all accented chars with regular ones
 */
function replaceAccents(str) {
  // Verifies if the String has accents and replace them
  if (str.search(/[\xC0-\xFF]/g) > -1) {
    str = str
      .replace(/[\xC0-\xC5]/g, "A")
      .replace(/[\xC6]/g, "AE")
      .replace(/[\xC7]/g, "C")
      .replace(/[\xC8-\xCB]/g, "E")
      .replace(/[\xCC-\xCF]/g, "I")
      .replace(/[\xD0]/g, "D")
      .replace(/[\xD1]/g, "N")
      .replace(/[\xD2-\xD6\xD8]/g, "O")
      .replace(/[\xD9-\xDC]/g, "U")
      .replace(/[\xDD]/g, "Y")
      .replace(/[\xDE]/g, "P")
      .replace(/[\xE0-\xE5]/g, "a")
      .replace(/[\xE6]/g, "ae")
      .replace(/[\xE7]/g, "c")
      .replace(/[\xE8-\xEB]/g, "e")
      .replace(/[\xEC-\xEF]/g, "i")
      .replace(/[\xF1]/g, "n")
      .replace(/[\xF2-\xF6\xF8]/g, "o")
      .replace(/[\xF9-\xFC]/g, "u")
      .replace(/[\xFE]/g, "p")
      .replace(/[\xFD\xFF]/g, "y");
  }

  return str;
}

Source: https://gist.github.com/jonlabelle/5375315

To remove all non-digit characters you can use

replaceAll("\\D+",""); \\ \D is negation of \d (where \d represents digit) 

If you want to remove only spaces, ( and ) you can define your own character class like

replaceAll("[\\s()]+","");

Anyway your problem was caused by fact that some of characters in regex are special. Among them there is ( which can represent for instance start of the group. Similarly ) can represent end of the group.

To make such special characters literals you need to escape them. You can do it many ways

• "\\(" - standard escaping in regex
• "[(]" - escaping using character class
• "\\Q(\\E" - \Q and \E create quote - which means that regex metacharacters in this area should be treated as simple literals
• Pattern.quote("(")) - this method uses Pattern.LITERAL flag inside regex compiler to point that metacharacters used in regex are simple literals without any special meaning

Just add the characters you want to remove in the character class:

replace(/[\s.;,?%0-9]/, '')

The OR operator (|) should work:

System.out.println(str.replaceAll("([^a-zA-Z]|\\s)+", " "));

Actually, the space doesn't have to be there at all:

System.out.println(str.replaceAll("[^a-zA-Z]+", " "));

Here is a regex to match a string of characters that are not a letters or numbers:

[^A-Za-z0-9]+

Here is the Python command to do a regex substitution:

re.sub('[^A-Za-z0-9]+', '', mystring)

You can do it specifying the characters you want to remove:

string = string.replace(/[&\/\\#,+()$~%.'":*?<>{}]/g, '');

Alternatively, to change all characters except numbers and letters, try:

string = string.replace(/[^a-zA-Z0-9]/g, '');