In the innermost loop, on line 17 of the function, you could change “c == arrClass.length” to “c == n”, since you already set n to the length of arrClass in the for statement.

In fact, you could simplify the compound if statement on line 17 by pre-computing the second operand of the comparison before the outermost loop (two lines added; three changed):


var arrClass = strClass.split(delim);
var n = arrClass.length;
var d = (delim == '|') ? 1 : n;
for (var i = 0, j = objColl.length; i arrObjClass.length) continue;
var c = 0;
comparisonLoop:
for (var k = 0, l = arrObjClass.length; k < l; k++) {
for (var m = 0; m < n; m++) {
if (arrClass[m] == arrObjClass[k]) c++;
if (c == d) {
arr.push(objColl[i]);
break comparisonLoop;
}
}
}
}


Also, if the user of getElementsByClassName accidentally puts more than one space between class names, as in…

var myObjColl = getElementsByClassName('one two');

… arrClass will contain a zero-length string, and the function will return an array with lots of extraneous, empty elements! Of course, you could argue that the user shouldn’t put more than one space between class names, but it wouldn’t be too difficult to make the function allow for this.

Really great ! Thanks for this !

var myObjColl = getElementsByClassName(’tt1′, ‘a’);
for (var i = 0, j = myObjColl.length; i

var c = getElementsByClassName('blah');

for(var i=0;j=c.length; i<j; i++){
  c[i].style.display = ‘none’;
}

anyone know if it is possible to create something that does
document.getElementsByClass(class).style.display =’none’;

basically the same as when using
document.getElementsById(id).style.display =’none’;

it would be a great help, if this would be possible

Thx in advance lotz

var cont = document.getElementById('container');
var myObjColl = getElementsByClassName('one two', 'a', cont);
for (var i = 0, j = myObjColl.length; i 

The argument ‘one two’ let you search for elements of class ‘one’ AND ‘two’ but I wanna search on a class name that has spaces in it.
My suggestion it to use something like ‘one&two’ as what you did in the “OR” case.