Javascript 获取节点的 XPath

Question

是否有在 Javascript 中返回 DOM 元素的 XPath 字符串？

Answer 1

我检查了此处提供的所有解决方案，但没有一个适用于 svg 元素（代码 getElementByXPath(getXPathForElement(elm)) === elm 返回 false 用于 svg 或 path 元素）

所以我将Touko's svg fix 添加到trincot's solution 并得到了这个代码：

function getXPathForElement(element) {
    const idx = (sib, name) => sib 
        ? idx(sib.previousElementSibling, name||sib.localName) + (sib.localName == name)
        : 1;
    const segs = elm => !elm || elm.nodeType !== 1 
        ? ['']
        : elm.id && document.getElementById(elm.id) === elm
            ? [`id("${elm.id}")`]
            : [...segs(elm.parentNode), elm instanceof HTMLElement
                ? `${elm.localName}[${idx(elm)}]`
                : `*[local-name() = "${elm.localName}"][${idx(elm)}]`];
    return segs(element).join('/');
}

不同之处在于，如果元素不是 HTMLElement 的实例（svg 是 SVGElement 但我决定不坚持只检查 svg），它返回 *[local-name() = "tag"][n] 而不是 tag[n]。

示例：

之前：
.../div[2]/div[2]/span[1]/svg[1]/path[1]

之后：
.../div[2]/div[2]/span[1]/*[local-name() = "svg"][1]/*[local-name() = "path"][1]

Answer 2

我已经调整了 algorithm Chromium uses 以从下面的 devtools 计算 XPath。

要使用这种书面形式，您可以致电Elements.DOMPath.xPath(<some DOM node>, false)。最后一个参数控制您是获得较短的“Copy XPath”（如果是true）还是“Copy full XPath”。

// Copyright 2018 The Chromium Authors. All rights reserved.
// Use of this source code is governed by a BSD-style license that can be
// found in the LICENSE file.

Elements = {};
Elements.DOMPath = {};

/**
 * @param {!Node} node
 * @param {boolean=} optimized
 * @return {string}
 */
Elements.DOMPath.xPath = function (node, optimized) {
    if (node.nodeType === Node.DOCUMENT_NODE) {
        return '/';
    }

    const steps = [];
    let contextNode = node;
    while (contextNode) {
        const step = Elements.DOMPath._xPathValue(contextNode, optimized);
        if (!step) {
            break;
        }  // Error - bail out early.
        steps.push(step);
        if (step.optimized) {
            break;
        }
        contextNode = contextNode.parentNode;
    }

    steps.reverse();
    return (steps.length && steps[0].optimized ? '' : '/') + steps.join('/');
};

/**
 * @param {!Node} node
 * @param {boolean=} optimized
 * @return {?Elements.DOMPath.Step}
 */
Elements.DOMPath._xPathValue = function (node, optimized) {
    let ownValue;
    const ownIndex = Elements.DOMPath._xPathIndex(node);
    if (ownIndex === -1) {
        return null;
    }  // Error.

    switch (node.nodeType) {
        case Node.ELEMENT_NODE:
            if (optimized && node.getAttribute('id')) {
                return new Elements.DOMPath.Step('//*[@id="' + node.getAttribute('id') + '"]', true);
            }
            ownValue = node.localName;
            break;
        case Node.ATTRIBUTE_NODE:
            ownValue = '@' + node.nodeName;
            break;
        case Node.TEXT_NODE:
        case Node.CDATA_SECTION_NODE:
            ownValue = 'text()';
            break;
        case Node.PROCESSING_INSTRUCTION_NODE:
            ownValue = 'processing-instruction()';
            break;
        case Node.COMMENT_NODE:
            ownValue = 'comment()';
            break;
        case Node.DOCUMENT_NODE:
            ownValue = '';
            break;
        default:
            ownValue = '';
            break;
    }

    if (ownIndex > 0) {
        ownValue += '[' + ownIndex + ']';
    }

    return new Elements.DOMPath.Step(ownValue, node.nodeType === Node.DOCUMENT_NODE);
};

/**
 * @param {!Node} node
 * @return {number}
 */
Elements.DOMPath._xPathIndex = function (node) {
    // Returns -1 in case of error, 0 if no siblings matching the same expression,
    // <XPath index among the same expression-matching sibling nodes> otherwise.
    function areNodesSimilar(left, right) {
        if (left === right) {
            return true;
        }

        if (left.nodeType === Node.ELEMENT_NODE && right.nodeType === Node.ELEMENT_NODE) {
            return left.localName === right.localName;
        }

        if (left.nodeType === right.nodeType) {
            return true;
        }

        // XPath treats CDATA as text nodes.
        const leftType = left.nodeType === Node.CDATA_SECTION_NODE ? Node.TEXT_NODE : left.nodeType;
        const rightType = right.nodeType === Node.CDATA_SECTION_NODE ? Node.TEXT_NODE : right.nodeType;
        return leftType === rightType;
    }

    const siblings = node.parentNode ? node.parentNode.children : null;
    if (!siblings) {
        return 0;
    }  // Root node - no siblings.
    let hasSameNamedElements;
    for (let i = 0; i < siblings.length; ++i) {
        if (areNodesSimilar(node, siblings[i]) && siblings[i] !== node) {
            hasSameNamedElements = true;
            break;
        }
    }
    if (!hasSameNamedElements) {
        return 0;
    }
    let ownIndex = 1;  // XPath indices start with 1.
    for (let i = 0; i < siblings.length; ++i) {
        if (areNodesSimilar(node, siblings[i])) {
            if (siblings[i] === node) {
                return ownIndex;
            }
            ++ownIndex;
        }
    }
    return -1;  // An error occurred: |node| not found in parent's children.
};

/**
 * @unrestricted
 */
Elements.DOMPath.Step = class {
    /**
     * @param {string} value
     * @param {boolean} optimized
     */
    constructor(value, optimized) {
        this.value = value;
        this.optimized = optimized || false;
    }

    /**
     * @override
     * @return {string}
     */
    toString() {
        return this.value;
    }
};

Answer 3

这是一个函数式编程风格的ES6函数：

function getXPathForElement(element) {
    const idx = (sib, name) => sib 
        ? idx(sib.previousElementSibling, name||sib.localName) + (sib.localName == name)
        : 1;
    const segs = elm => !elm || elm.nodeType !== 1 
        ? ['']
        : elm.id && document.getElementById(elm.id) === elm
            ? [`id("${elm.id}")`]
            : [...segs(elm.parentNode), `${elm.localName.toLowerCase()}[${idx(elm)}]`];
    return segs(element).join('/');
}

function getElementByXPath(path) { 
    return (new XPathEvaluator()) 
        .evaluate(path, document.documentElement, null, 
                        XPathResult.FIRST_ORDERED_NODE_TYPE, null) 
        .singleNodeValue; 
} 

// Demo:
const li = document.querySelector('li:nth-child(2)');
const path = getXPathForElement(li);
console.log(path);
console.log(li === getElementByXPath(path)); // true

<div>
    <table id="start"></table>
    <div>
        <ul><li>option</ul></ul> 
        <span>title</span>
        <ul>
            <li>abc</li>
            <li>select this</li>
        </ul>
    </div>
</div>

它将使用id 选择器，除非该元素不是第一个具有该ID 的元素。不使用类选择器，因为在交互式网页中，类可能经常更改。

Answer 4

只需在函数getXPathOfElement 中传递元素，您就会得到Xpath。

function getXPathOfElement(elt)
{
     var path = "";
     for (; elt && elt.nodeType == 1; elt = elt.parentNode)
     {
    idx = getElementIdx(elt);
    xname = elt.tagName;
    if (idx > 1) xname += "[" + idx + "]";
    path = "/" + xname + path;
     }

     return path;   
}
function getElementIdx(elt)
{
    var count = 1;
    for (var sib = elt.previousSibling; sib ; sib = sib.previousSibling)
    {
        if(sib.nodeType == 1 && sib.tagName == elt.tagName) count++
    }

    return count;
}

Answer 5

function getElementXPath (element) {
  if (!element) return null

  if (element.id) {
    return `//*[@id=${element.id}]`
  } else if (element.tagName === 'BODY') {
    return '/html/body'
  } else {
    const sameTagSiblings = Array.from(element.parentNode.childNodes)
      .filter(e => e.nodeName === element.nodeName)
    const idx = sameTagSiblings.indexOf(element)

    return getElementXPath(element.parentNode) +
      '/' +
      element.tagName.toLowerCase() +
      (sameTagSiblings.length > 1 ? `[${idx + 1}]` : '')
  }
}

console.log(getElementXPath(document.querySelector('#a div')))

<div id="a">
 <div>def</div>
</div>

Answer 6

MDN上的函数getXPathForElement给出了类似的解决方案：

function getXPathForElement(el, xml) {
    var xpath = '';
    var pos, tempitem2;

    while(el !== xml.documentElement) {     
        pos = 0;
        tempitem2 = el;
        while(tempitem2) {
            if (tempitem2.nodeType === 1 && tempitem2.nodeName === el.nodeName) { // If it is ELEMENT_NODE of the same name
                pos += 1;
            }
            tempitem2 = tempitem2.previousSibling;
        }

        xpath = "*[name()='"+el.nodeName+"' and namespace-uri()='"+(el.namespaceURI===null?'':el.namespaceURI)+"']["+pos+']'+'/'+xpath;

        el = el.parentNode;
    }
    xpath = '/*'+"[name()='"+xml.documentElement.nodeName+"' and namespace-uri()='"+(el.namespaceURI===null?'':el.namespaceURI)+"']"+'/'+xpath;
    xpath = xpath.replace(/\/$/, '');
    return xpath;
}

XMLSerializer 也值得一试。

Answer 7

我从另一个例子中重构了这个。它将尝试检查或确定是否存在唯一 id，如果是，则使用这种情况来缩短表达式。

function createXPathFromElement(elm) { 
    var allNodes = document.getElementsByTagName('*'); 
    for (var segs = []; elm && elm.nodeType == 1; elm = elm.parentNode) 
    { 
        if (elm.hasAttribute('id')) { 
                var uniqueIdCount = 0; 
                for (var n=0;n < allNodes.length;n++) { 
                    if (allNodes[n].hasAttribute('id') && allNodes[n].id == elm.id) uniqueIdCount++; 
                    if (uniqueIdCount > 1) break; 
                }; 
                if ( uniqueIdCount == 1) { 
                    segs.unshift('id("' + elm.getAttribute('id') + '")'); 
                    return segs.join('/'); 
                } else { 
                    segs.unshift(elm.localName.toLowerCase() + '[@id="' + elm.getAttribute('id') + '"]'); 
                } 
        } else if (elm.hasAttribute('class')) { 
            segs.unshift(elm.localName.toLowerCase() + '[@class="' + elm.getAttribute('class') + '"]'); 
        } else { 
            for (i = 1, sib = elm.previousSibling; sib; sib = sib.previousSibling) { 
                if (sib.localName == elm.localName)  i++; }; 
                segs.unshift(elm.localName.toLowerCase() + '[' + i + ']'); 
        }; 
    }; 
    return segs.length ? '/' + segs.join('/') : null; 
}; 

function lookupElementByXPath(path) { 
    var evaluator = new XPathEvaluator(); 
    var result = evaluator.evaluate(path, document.documentElement, null,XPathResult.FIRST_ORDERED_NODE_TYPE, null); 
    return  result.singleNodeValue; 
} 

Answer 8

节点没有唯一的 XPath，因此您必须确定构建路径的最合适方式是什么。在可用的情况下使用 ID？数字在文档中的位置？相对于其他元素的位置？

请参阅this answer 中的getPathTo() 了解一种可能的方法。