Prototype ¬ Detail

author: dark sylkro [+], Submitted: 12.28.01 5a • Last Edit: 10.01.02 12p

Array.prototype.crossRef = function(arr1, arr2, func) {
	var arr1, arr2, func, common = new Array();
	this.length = 0;
	for (var i=0; i<arr1.length; i++) {
		for (var j=0; j<arr2.length; j++) {
			if (arr1[i] == arr2[j]) {
				common.push(arr1[i]);
			}
		}
	}
	if (func or func == undefined) {
		for (var i=0; i<common.length; i++) {
			this.push(common[i]);
		}
	} else {
		var negObj = new Object(), whole = arr1.concat(arr2);
		//
		for (var i=0; i<whole.length; i++) {
			negObj[whole[i]] = true;
		}
		for (var i=0; i<common.length; i++) {
			negObj[common[i]] = false;
		}
		//
		for (var iter in negObj) {
			if (negObj[iter]) {
				this.push(iter);
			}
		}
	}
	return(this);
}

usage

analyse 2 given arrays and either return an array of
elements which they have in common, or an array of elements
they don't have in common.

the returned result is automatically palced in the third
array and does not affect the original 2 arrays.  Finally,
it will not filter out duplicates, there's a method on this
site that does that already if you're interested.

So:
array1 = new Array("Mon", "Tue", "Wed", "Thu");
array2 = new Array("Thu", "Fri", "Sat", "Sun");
//
array3.crossRef(array1, array2, true);
array3.crossRef(array1, array2);
//
// the above statements return common elements into array3:
// "Thu"

array3.crossRef(array1, array2, false);
//
// this however returns an array of the elements that aren't common
// "Mon", "Tue", "Wed", "Fri", "Sat", "Sun"


thanks to Gary Ma for finding a bug in the inverted function