查找对象的超级嵌套数组的位置

Question

我试图找到确切的位置并访问对象的超级嵌套数组的所有属性。

我正在努力创建一个函数，如果我给定索引号作为输入参数，则应该给我它在数组中的位置，并返回所有属性。

这是对象的示例数组

我也同意ES6及以上版本的解决方案

{
	"name": "branch 1",
	"index": 1,
	"children": [{
			"name": "sub child 1",
			"index": 2,
			"children": [{
				"name": "subx2 child 1",
				"index": 3,
				"children": [{
						"name": "subx3 child 1",
						"index": 4,
						"children": [{
								"name": "subx4 child 1",
								"index": 21
							},
							{
								"name": "subx4 child 2",
								"index": 18
							}
						]
					},
					{
						"name": "subx3 child 2",
						"index": 6,
						"children": [{
								"name": "subx4 child 1",
								"index": 7
							},
							{
								"name": "subx4 child 2",
								"index": 21
							}
						]

					},
					{
						"name": "subx3 child 3",
						"index": 22
					}
				]
			}]
		},
		{
			"name": "sub child 2",
			"index": 28
		}
	]
}

是的，我知道这个json对象足够吓人，无法花费时间来解决。任何帮助都非常有用。

例如，如果我的函数名称是 findChildIndex（22），则应该返回类似 x.children [0] .children [0] .children [2]

谢谢！

Answer 1

您可以递归地收集children数组中通往目标索引的索引：

function findIndexNested(data, index) {
    if (data.index === index) return [];
    let result;
    const i = (data.children || []).findIndex(child => {
        return result = findIndexNested(child, index)
    });
    if (result) return [i, ...result];
}
function findByPath(data, path) {
    for (let i of path) data = data.children[i];
    return data
}
// Sample data
const data = {"name": "branch 1","index": 1,"children": [{"name": "sub child 1","index": 2,"children": [{"name": "subx2 child 1","index": 3,"children": [{"name": "subx3 child 1","index": 4,"children": [{"name": "subx4 child 1","index": 21},{"name": "subx4 child 2","index": 18}]},{"name": "subx3 child 2","index": 6,"children": [{"name": "subx4 child 1","index": 7},{"name": "subx4 child 2","index": 21}]},{"name": "subx3 child 3","index": 22}]}]},{"name": "sub child 2","index": 28}]}
const index = 22
const result = findIndexNested(data, index);
console.log("Found index " + index + " via these child indexes: " + result);
console.log("The object is", findByPath(data, result));

Answer 2

您可以使用递归并检查元素children是否存在，可以使用for循环遍历所有children并递归地应用每个子元素的功能

const obj = {
	"name": "branch 1",
	"index": 1,
	"children": [{
			"name": "sub child 1",
			"index": 2,
			"children": [{
				"name": "subx2 child 1",
				"index": 3,
				"children": [{
						"name": "subx3 child 1",
						"index": 4,
						"children": [{
								"name": "subx4 child 1",
								"index": 21
							},
							{
								"name": "subx4 child 2",
								"index": 18
							}
						]
					},
					{
						"name": "subx3 child 2",
						"index": 6,
						"children": [{
								"name": "subx4 child 1",
								"index": 7
							},
							{
								"name": "subx4 child 2",
								"index": 21
							}
						]

					},
					{
						"name": "subx3 child 3",
						"index": 22
					}
				]
			}]
		},
		{
			"name": "sub child 2",
			"index": 28
		}
	]
}

function find(obj,index){
  if(obj.children){
    for(let i = 0;i<obj.children.length;i++){
      let x = find(obj.children[i],index);
      if(x) return {...x,pos:i};
    }

  }
  return obj.index === index ? obj : false;
}

console.log(find(obj,21))

Answer 3

如果我正确回答了您的问题，则可以执行以下操作：

const func=(obj,index, nested=0)=>{
    return Obj.index===index ? {obj, nested} : func(obj.children,index, nested+1)
}