javascript - 用于访问嵌套数组中元素的递归函数和 Map

Question

情况

• 我有一个嵌套数组，并且有一个要搜索的 ID。类似于“A_02_02_01_03”
• 数组中的每个元素都有一个称为“子元素”的元素，即数组
• 当我在第 4 层搜索时，我的方法会变得相当长。

数组示例

var tree= [
  {
    "name": "i2",
    "children": [
      {
        "name": "d1",
        "children": [],
        "id": "DW_02_01",
        "beschreibung": "",
        "table": []
      },
      {
        "name": "d2",
        "children": [
          {
            "name": "e1",
            "children": [
              {
                "name": "a1",
                "children": [],
                "id": "A_02_02_01_01",
                "beschreibung": "",
                "table": []
              },
              {
                "name": "a2",
                "children": [],
                "id": "A_02_02_01_02",
                "beschreibung": "",
                "table": []
              },
              {
                "name": "a3",
                "children": [],
                "id": "A_02_02_01_03",
                "beschreibung": "",
                "table": []
              }`enter code here`
            ],
            "id": "E_02_02_01",
            "beschreibung": "",
            "table": []
          },
          {
            "name": "e2",
            "children": [],
            "id": "E_02_02_02",
            "beschreibung": "",
            "table": []
          }
        ],
        "id": "DW_02_02",
        "beschreibung": "",
        "table": []
      },
      {
        "name": "d3",
        "children": [],
        "id": "DW_02_03",
        "beschreibung": "",
        "table": []
      }
    ],
    "id": "IW_02",
    "beschreibung": "",
    "table": []
  },
  {
    "name": "i3",
    "children": [],
    "id": "IW_03",
    "beschreibung": "",
    "table": []
  }
]

构造 ID

var daIW = "IW_02";
var daDW = "DW_02_02;
var daE = "E_02_02_01;
var daA = "A_02_02_01_03";

获取我的所有索引

var iw_index = tree.findIndex(element => element.id == daIW);
var dw_index = tree[iw_index]["children"].findIndex(element => element.id == daDW);
var e_index = tree[iw_index]["children"][dw_index]["children"].findIndex(element => element.id == daE);
var a_index = tree[iw_index]["children"][dw_index]["children"][e_index]["children"].findIndex(element => element.id == daA);

访问我的元素

var elementName = tree[iw_index]["children"][dw_index]["children"][e_index]["children"][a_index].name;

问题

是否有更短的方法来访问最深的元素“A_02_02_01_03”然后搜索每个索引？

Answer 1

您可能希望递归首先深入搜索树:

function search(array = [], id){
  for(const node of array){
    if(node.id === id) return node;

    const sub = search(node.children, id);
    if(sub) return sub;
  }
}

所以你可以这样做:

const result = search(tree, "A_02_02_01_03");

<小时/>

如果你想查找多个项目，最好建立一个存储所有 id/node 对的哈希表，这样查找速度非常快:

function createLookup(array, hash = new Map){
  for(const node of array){
    hash.set(node.id, node);
    createLookup(node.children, hash);
  }
  return hash;
}

所以你可以这样做:

const hash = createLookup(tree);
const result = hash.get("A_02_02_01_03");