如何选择包含全部特定文本后代元素的最底层祖先元素？

你的需求本质是找所有含"SOMETHING"元素的最深共同祖先（Lowest Common Ancestor, LCA），以下是两种可行方案：

原生JavaScript方案 #

先定位所有目标元素，再从它们的祖先中筛选出最深的共同节点：

// 1. 获取所有文本为"SOMETHING"的元素
const targetElements = Array.from(document.querySelectorAll('div'))
  .filter(el => el.textContent.trim() === 'SOMETHING');

// 2. 实现找最深共同祖先的函数
function getDeepestCommonAncestor(elements) {
  if (elements.length === 0) return null;

  // 收集第一个元素的所有祖先（包括自身）
  let ancestorSet = new Set();
  let current = elements[0];
  while (current) {
    ancestorSet.add(current);
    current = current.parentElement;
  }

  // 遍历其余元素，逐步缩小共同祖先范围
  for (let i = 1; i < elements.length; i++) {
    current = elements[i];
    let commonAncestors = new Set();
    while (current) {
      if (ancestorSet.has(current)) {
        commonAncestors.add(current);
      }
      current = current.parentElement;
    }

    if (commonAncestors.size === 0) return null;
    ancestorSet = commonAncestors;
  }

  // 从共同祖先中筛选出最深的那个
  let deepest = null;
  ancestorSet.forEach(el => {
    if (!deepest || el.contains(deepest)) {
      deepest = el;
    }
  });

  return deepest;
}

// 执行并获取结果
const result = getDeepestCommonAncestor(targetElements);
console.log(result); // 对应示例中的div C

XPath方案 #

用一条XPath表达式直接定位目标元素，核心逻辑是：找到包含所有"SOMETHING"元素，且其子元素都无法包含所有目标元素的节点（即最深的共同祖先）。

//*[
  count(.//*[contains(text(),"SOMETHING")]) = count(//*[contains(text(),"SOMETHING")]) 
  and not(
    ./child::*[count(.//*[contains(text(),"SOMETHING")]) = count(//*[contains(text(),"SOMETHING")])]
  )
]

在JS中使用该XPath的方式：

const xpathResult = document.evaluate(
  '//*[count(.//*[contains(text(),"SOMETHING")]) = count(//*[contains(text(),"SOMETHING")]) and not(./child::*[count(.//*[contains(text(),"SOMETHING")]) = count(//*[contains(text(),"SOMETHING")])])]',
  document,
  null,
  XPathResult.FIRST_ORDERED_NODE_TYPE,
  null
);
const result = xpathResult.singleNodeValue;
console.log(result);

内容的提问来源于stack exchange，提问作者Damzaky