'Distance' between elements in DOM (generation depth etc.)

Question

I was looking for a function that would return the 'distance' between elements in the DOM in terms of ancestors, siblings and descendants. For example, suppose I had:

<div id="div1">
    <div id="div5"></div>
</div>
<div id="div2">
    <div id="div6">
        <div id="div9"></div>
    </div>
    <div id="div7"></div>
</div>
<div id="div3"></div>
<div id="div4">
    <div id="div8">
        <div id="div10"></div>
    </div>
</div>

then I'd want a function that would return the distance between #div5 and #div10 as something like:

{
    up: 1,
    across: 3,
    down: 2
}

Since to get from #div5 to #div10 you have to go up one generation, forward 3 siblings (to #div4) and then down 2 generations. Similarly, #div9 to #div1 would return:

{
    up: 2,
    across: -1,
    down: 0
}

for going up two generations, and then back one sibling.

I've already got a function that does this (which I'll include as an answer below) so I'm including it here because a) I thought someone else might find it useful; and b) maybe someone else has a better way of doing it.

Answer 1

Ok, here's what I have for it. I've hopefully explained it well enough in the code comments:

function DOMdistance(elem1,elem2) {

    if (elem1 === elem2) {
        return {
            up: 0,
            across: 0,
            down: 0
        };
    }

    var parents1 = [elem1],
        parents2 = [elem2],
        gens = 1,
        sibs = 0,
        sibElem;

    // searches up the DOM from elem1 to the body, stopping and 
    // returning if it finds elem2 as a direct ancestor
    while (elem1 !== document.body) {
        elem1 = elem1.parentNode;
        if (elem1 === elem2) {
            return {
                up: parents1.length,
                across: 0,
                down: 0
            };
        }
        parents1.unshift(elem1);
    }

    // reset value of elem1 for use in the while loop that follows:
    elem1 = parents1[parents1.length - 1];

    // searches up the DOM from elem2 to the body, stopping and 
    // returning if it finds elem1 as a direct ancestor
    while (elem2 !== document.body) {
        elem2 = elem2.parentNode;
        if (elem2 === elem1) {
            return {
                up: 0,
                across: 0,
                down: parents2.length
            };
        }
        parents2.unshift(elem2);
    }

    // finds generation depth from body of first generation of ancestors 
    // of elem1 and elem2 that aren't common to both
    while (parents1[gens] === parents2[gens]) {
        gens++;
    }

    sibElem = parents1[gens];

    // searches forward across siblings from the earliest non-common ancestor
    // of elem1, looking for earliest non-common ancestor of elem2
    while (sibElem) {
        sibElem = sibElem.nextSibling;
        if (sibElem && sibElem.tagName) {
            sibs++;
            if (sibElem === parents2[gens]) {
                return {
                    up: parents1.length - gens - 1,
                    across: sibs,
                    down: parents2.length - gens - 1
                };
            }
        }
    }

    sibs = 0;
    sibElem = parents1[gens];

    // searches backward across siblings from the earliest non-common ancestor 
    // of elem1, looking for earliest non-common ancestor of elem2
    while (sibElem) {
        sibElem = sibElem.previousSibling;
        if (sibElem && sibElem.tagName) {
            sibs--;
            if (sibElem === parents2[gens]) {
                return {
                    up: parents1.length - gens - 1,
                    across: sibs,
                    down: parents2.length - gens - 1
                };
            }
        }
    }

}

So, for example, getting the 'distance' from #div5 to #div10 in the DOM described in the question would use something like:

var divOne = document.getElementById('div5'),
    divTwo = document.getElementById('div10'),
    distance = DOMdistance(divOne, divTwo);

and therefore distance would be:

{
    up: 1,
    across: 3,
    down: 2
}

demo: http://jsfiddle.net/x58Ga/