Emulating SQL LIKE in JavaScript

I was looking for an answer the same question and came up with this after reading Kip's reply:

String.prototype.like = function(search) {
    if (typeof search !== 'string' || this === null) {return false; }
    // Remove special chars
    search = search.replace(new RegExp("([\\.\\\\\\+\\*\\?\\[\\^\\]\\$\\(\\)\\{\\}\\=\\!\\<\\>\\|\\:\\-])", "g"), "\\$1");
    // Replace % and _ with equivalent regex
    search = search.replace(/%/g, '.*').replace(/_/g, '.');
    // Check matches
    return RegExp('^' + search + '$', 'gi').test(this);
}

You can then use it as follows (note that it ignores UPPER/lower case):

var url = 'http://www.mydomain.com/page1.aspx';
console.log(url.like('%mydomain.com/page_.asp%')); // true

NOTE 29/11/2013: Updated with RegExp.test() performance improvement as per Lucios comment below.

What you have will work as long as you first escape the regex characters in your pattern. Below is one example from Simon Willison’s blog:

RegExp.escape = function(text) {
  if (!arguments.callee.sRE) {
    var specials = [
      '/', '.', '*', '+', '?', '|',
      '(', ')', '[', ']', '{', '}', '\\'
    ];
    arguments.callee.sRE = new RegExp(
      '(\\' + specials.join('|\\') + ')', 'g'
    );
  }
  return text.replace(arguments.callee.sRE, '\\$1');
}

You could then implement your code as:

likeExpr = RegExp.escape(likeExpr);
var match = new RegEx(likeExpr.replace("%", ".*").replace("_", ".")).exec(str) != null;

Here's a function I use, based on PHP's preg_quote function:

function regex_quote(str) {
  return str.replace(new RegExp("([\\.\\\\\\+\\*\\?\\[\\^\\]\\$\\(\\)\\{\\}\\=\\!\\<\\>\\|\\:\\-])", "g"), "\\$1");
}

So your line would now be:

var match = new RegEx(regex_quote(likeExpr).replace("%", ".*").replace("_", ".")).exec(str) != null;