如何从嵌套的对象数组中获取每个父级的值

所以我有多个对象数组，每个对象都包含一个子对象。

e.g

const data = [
    {
        id: 1,
        name: 'parent 1',
        children: [
            {
                id: 'c1',
                name: 'child 1',
                children: [
                    {
                        id: 'g1',
                        name: 'grand 1',
                        children: [],
                    },
                ],
            },
        ],
    },
    {
        id: 2,
        name: 'parent 2',
        children: [
            {
                id: 2,
                name: 'c1',
                children: [],
            },
        ],
    },
    { id: 3, name: 'parent 3', children: [] },
];

我想要发生的是，如果我正在搜索的 ID 是 'g1'，我会得到结果

const result = ['parent 1', 'c1', 'grand 1']

循环只会停止并获取它所经过的所有名称，直到满足条件（在本例中为 id）

当前方法已完成

/**
 * Details
 * @param id the value you are searching for
 * @param items nested array of object that has child
 * @param key name of the value you are looking for
 * @returns string of array that matches the id
 * @example ['parent 1', 'c1', 'grand 1']
 */
export function findAll(id: string, items: any, key: string): string[] {
  let i = 0;
  let found;
  let result = [];

  for (; i < items.length; i++) {
    if (items[i].id === id) {
      result.push(items[i][key]);
    } else if (_.isArray(items[i].children)) {
      found = findAll(id, items[i].children, key);
      if (found.length) {
        result = result.concat(found);
      }
    }
  }
  return result;
}

我编写的这段迭代代码可能会对您有所帮助。它基本上遍历存储从顶层到所需 id 的路径的结构：

function getPath(obj, id) {
    // We need to store path
    // Start stack with root nodes
    let stack = obj.map(item => ({path: [item.name], currObj: item}));
    while (stack.length) {
        const {path, currObj} = stack.pop()
        if (currObj.id === id) {
            return path;
        } else if (currObj.children?.length) {
            stack = stack.concat(currObj.children.map(item => ({path: path.concat(item.name), currObj: item})));
        }
    }
    return null; // if id does not exists
}

此代码假设您的结构是正确的并且没有丢失任何部分（除了可以为空的子项）。 顺便问一下，你的答案正确吗？我想路径应该是：["parent 1", "child 1", "grand 1"]