Question

我正在寻找一个函数，它可以在祖先，兄弟和后代方面返回DOM中元素之间的“距离”。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>

然后我想要一个能够返回#div5和#div10之间距离的函数，如：

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

从#div5到#div10，你必须上升一代，转发3个兄弟姐妹（到#div4）然后再向下2代。同样，#div9到#div1会返回：

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

上升两代，然后回到一个兄弟。

我已经有了这样做的功能（我将在下面作为答案包括在内）所以我在这里包括它因为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，这就是我拥有的东西。我希望在代码评论中能够很好地解释它：

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，在问题中描述的DOM中获取#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distance将是：

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

演示：http://jsfiddle.net/x58Ga/

DOM中元素之间的“距离”（生成深度等）

1 个答案: