javascript - DOM 中元素之间的“距离”(生成深度等)

Question

我正在寻找一个函数，它可以返回 DOM 中元素之间的“距离”(以祖先、 sibling 和后代为单位)。例如，假设我有:

<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>

然后我想要一个函数来返回 #div5 和 #div10 之间的距离，如下所示:

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

由于要从 #div5 到 #div10，您必须向上一代，转发 3 个 sibling (到 #div4)并且然后下降了2代。同样，#div9 到 #div1 将返回:

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

向上两代，然后返回一个 sibling 。

我已经有了一个可以执行此操作的函数(我将在下面将其作为答案包含在内)，因此我将其包含在此处，因为 a)我认为其他人可能会发现它有用； b) 也许其他人有更好的方法。

Answer 1

好的，这就是我所拥有的。我希望在代码注释中对其进行了足够好的解释:

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
                };
            }
        }
    }

}

因此，例如，获取问题中描述的 DOM 中从 #div5 到 #div10 的“距离”将使用如下内容:

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

因此距离将是:

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

演示:http://jsfiddle.net/x58Ga/