Question

让我先向您展示代码，然后再解释我的需求。我想向您展示输入数组的结构和所需的结果：

[
  {
    link: "someurl",
    name: "Foo",
    subCats: [
      {
        link: "anotherurl",
        name: "Bar",
        subCats: [
          {
            link: "anotherurl",
            subCats: [
              {
                link: "onemorekink"
                name: "Prod",
              }
            ]
          }
        ]
      },
      {
        link: "someurll",
        name: "Fuzzy",
        subCats: [
          {
            link: "besturiever",
            name: "SomeName",
            subCats: [
              {
                link: "onemore",
                name: "Aloc",
                subCats: [
                  {
                    link: "anotherlink"
                    name: "Final",
                  }
                ]
              }
            ]
          }
        ]
      }
    ]
  }
]

我需要得到什么？

{
  link: "onemorekink"
  name: "Prod",
},
{
  link: "anotherlink"
  name: "Final",
}

希望您能明白。基本上，我需要以某种方式获取不包含子项subCats的最后一个subCats元素，并将其追加到结果数组中。我尝试使用Lodash，因为它非常适合数组/对象操作。谢谢您的帮助。

Answer 1

编写一种无需lodash即可为您完成此操作的方法很简单

function findLastLeafs(items, key) {
  let results = [];
  items.forEach((item, index) => {
    if (item[key] && item[key].length) {
      results = results.concat(findLastLeafs(item[key], key));
    } else if (index === items.length - 1) {
      results.push(item);
    }
  })
  return results;
}

const result = findLastLeafs(data, 'subCats');

Answer 2

尝试一下

let b=[]; // result here
let f = x => x.map( y=> y.subCats ? f(y.subCats) : b.push(y) );
f(a);

let a = [
  {
    link: "someurl",
    name: "Foo",
    subCats: [
      {
        link: "anotherurl",
        name: "Bar",
        subCats: [
          {
            link: "anotherurl",
            subCats: [
              {
                link: "onemorekink",
                name: "Prod",
              }
            ]
          }
        ]
      },
      {
        link: "someurll",
        name: "Fuzzy",
        subCats: [
          {
            link: "besturiever",
            name: "SomeName",
            subCats: [
              {
                link: "onemore",
                name: "Aloc",
                subCats: [
                  {
                    link: "anotherlink",
                    name: "Final",
                  }
                ]
              }
            ]
          }
        ]
      }
    ]
  }
];

let b=[];
let f = x => x.map( y=> y.subCats ? f(y.subCats) : b.push(y) );
f(a);

console.log(b);

Answer 3

挺直的，不需要lodash。递归函数..

function process(data){
        for(let a of data){
            if(a.subCats){
                process(a.subCats);
            }else{
                console.log(a)
            }
        }
    }

Lodash从混合数组和对象中获取叶子

3 个答案: