如何用JavaScript提取嵌套文件夹对象数组的有效路径？

问题描述 #

给定如下JSON格式的嵌套文件夹结构：

[{ 
  "name": "Knowledge Base",
  "files": [
    {
      "name": "Documents",
      "files": [
        {
          "name": "Quarterly Results"
        }
      ]
    },
    {
      "name": "Favourites",
      "files": [
        {
          "name": "Brawl Stars",
          "files": [
            {
              "name": "NS dying in 5 seconds"
            },
            {
              "name": "Josiah raping NS"
            }
          ]
        },
        {
          "name": "Coding",
          "files": [
            {
              "name": "Coding is so fun"
            },
            {
              "name": "I love svelte",
              "files": [
                {
                  "name": "REPL"
                },
                {
                  "name": "oh nooo"
                }
              ]
            }
          ]
        },
        {
          "name": "Favourites 1"
        },
        {
          "name": "Favourites 2"
        },
        {
          "name": "Favourites 3"
        }
      ]
    },
    {
      "name": "Knowledge Base 1"
    }
  ]
}]

需要提取所有包含子文件夹的层级路径（忽略无下级文件夹的文件），期望输出：

• Knowledge Base > Documents
• Knowledge Base > Favourites > Brawl Stars
• Knowledge Base > Favourites > Coding
• Knowledge Base > Favourites > Coding > I love svelte

递归实现方案 #

用递归遍历每个节点，核心逻辑是：只有当节点存在files子数组且不为空时，才记录该节点的完整路径，再递归处理其子节点。

以JavaScript为例，代码实现如下：

// 输入的文件夹结构
const folderStructure = [{ 
  "name": "Knowledge Base",
  "files": [
    {
      "name": "Documents",
      "files": [
        {
          "name": "Quarterly Results"
        }
      ]
    },
    {
      "name": "Favourites",
      "files": [
        {
          "name": "Brawl Stars",
          "files": [
            {
              "name": "NS dying in 5 seconds"
            },
            {
              "name": "Josiah raping NS"
            }
          ]
        },
        {
          "name": "Coding",
          "files": [
            {
              "name": "Coding is so fun"
            },
            {
              "name": "I love svelte",
              "files": [
                {
                  "name": "REPL"
                },
                {
                  "name": "oh nooo"
                }
              ]
            }
          ]
        },
        {
          "name": "Favourites 1"
        },
        {
          "name": "Favourites 2"
        },
        {
          "name": "Favourites 3"
        }
      ]
    },
    {
      "name": "Knowledge Base 1"
    }
  ]
}];

// 存储结果的数组
const resultPaths = [];

// 递归函数
function traverseFolders(node, currentPath = '') {
  // 拼接当前节点的路径
  const newPath = currentPath ? `${currentPath} > ${node.name}` : node.name;
  
  // 检查当前节点是否有子文件夹（files存在且不为空）
  if (node.files && node.files.length > 0) {
    // 记录该路径
    resultPaths.push(newPath);
    // 递归遍历所有子节点
    node.files.forEach(child => {
      traverseFolders(child, newPath);
    });
  }
}

// 启动递归遍历根节点
folderStructure.forEach(root => traverseFolders(root));

// 输出结果
console.log(resultPaths);
// 输出内容：
// [
//   "Knowledge Base > Documents",
//   "Knowledge Base > Favourites > Brawl Stars",
//   "Knowledge Base > Favourites > Coding",
//   "Knowledge Base > Favourites > Coding > I love svelte"
// ]

代码说明 #

1. 递归函数traverseFolders：接收当前节点和当前路径前缀，每次调用先拼接出当前节点的完整路径。
2. 判断条件：只有当节点的files属性存在且数组不为空时，才将当前路径加入结果集，因为这说明该节点有子文件夹。
3. 递归遍历：对每个子节点递归调用函数，传递更新后的路径前缀，确保子节点的路径能正确拼接上级路径。

内容的提问来源于stack exchange，提问作者user20840709