How can I concatenate regex literals in JavaScript?

I don't quite agree with the "eval" option.

var xxx = /abcd/;
var yyy = /efgh/;
var zzz = new RegExp(eval(xxx)+eval(yyy));

will give "//abcd//efgh//" which is not the intended result.

Using source like

var zzz = new RegExp(xxx.source+yyy.source);

will give "/abcdefgh/" and that is correct.

Logicaly there is no need to EVALUATE, you know your EXPRESSION. You just need its SOURCE or how it is written not necessarely its value. As for the flags, you just need to use the optional argument of RegExp.

In my situation, I do run in the issue of ^ and $ being used in several expression I am trying to concatenate together! Those expressions are grammar filters used accross the program. Now I wan't to use some of them together to handle the case of PREPOSITIONS. I may have to "slice" the sources to remove the starting and ending ^( and/or )$ :) Cheers, Alex.

Just randomly concatenating regular expressions objects can have some adverse side effects. Use the RegExp.source instead:

var r1 = /abc/g;
var r2 = /def/;
var r3 = new RegExp(r1.source + r2.source, 
                   (r1.global ? 'g' : '') 
                   + (r1.ignoreCase ? 'i' : '') + 
                   (r1.multiline ? 'm' : ''));
console.log(r3);
var m = 'test that abcdef and abcdef has a match?'.match(r3);
console.log(m);
// m should contain 2 matches

This will also give you the ability to retain the regular expression flags from a previous RegExp using the standard RegExp flags.

jsFiddle

Here is how to create a regular expression without using the regular expression literal syntax. This lets you do arbitary string manipulation before it becomes a regular expression object:

var segment_part = "some bit of the regexp";
var pattern = new RegExp("some regex segment" + /*comment here */
              segment_part + /* that was defined just now */
              "another segment");

If you have two regular expression literals, you can in fact concatenate them using this technique:

var regex1 = /foo/g;
var regex2 = /bar/y;
var flags = (regex1.flags + regex2.flags).split("").sort().join("").replace(/(.)(?=.*\1)/g, "");
var regex3 = new RegExp(expression_one.source + expression_two.source, flags);
// regex3 is now /foobar/gy

It's just more wordy than just having expression one and two being literal strings instead of literal regular expressions.

Problem If the regexp contains back-matching groups like \1.

var r = /(a|b)\1/  // Matches aa, bb but nothing else.
var p = /(c|d)\1/   // Matches cc, dd but nothing else.

Then just contatenating the sources will not work. Indeed, the combination of the two is:

var rp = /(a|b)\1(c|d)\1/
rp.test("aadd") // Returns false

The solution: First we count the number of matching groups in the first regex, Then for each back-matching token in the second, we increment it by the number of matching groups.

function concatenate(r1, r2) {
  var count = function(r, str) {
    return str.match(r).length;
  }
  var numberGroups = /([^\\]|^)(?=\((?!\?:))/g; // Home-made regexp to count groups.
  var offset = count(numberGroups, r1.source);    
  var escapedMatch = /[\\](?:(\d+)|.)/g;        // Home-made regexp for escaped literals, greedy on numbers.
  var r2newSource = r2.source.replace(escapedMatch, function(match, number) { return number?"\\"+(number-0+offset):match; });
  return new RegExp(r1.source+r2newSource,
      (r1.global ? 'g' : '') 
      + (r1.ignoreCase ? 'i' : '')
      + (r1.multiline ? 'm' : ''));
}

Test:

var rp = concatenate(r, p) // returns  /(a|b)\1(c|d)\2/
rp.test("aadd") // Returns true