在Javascript

Question

我有一个格式如下的数组

 let array = [{id: 1, desc: 'd1', children:[{id:1.1, desc:'d1.1', children:[]}]}, 
                 {id:2, desc:'d2', children:[] }, 
                 {id:3, desc:'d3', children:[] }];

每个孩子与父元素同时出现的情况。我希望它将其转换为格式为{ [id]: {values} }的对象：

{
 1: { id: 1, desc: 'd1', children: {1.1: {id:1.1, desc:'d1.1'}},
 2: { id:2, desc:'d2' },
 3: { id:3, desc:'d3' }
}

我在很多方面都尝试过但没有成功。例如：

let obj = array.map(a => mapArrayToObj(a));

mapArrayToObj = (e) => { 
     let obj = {[e.id]: e };
     if(e.children.lenght > 0){
       e.children = e.children.map(c => mapArrayToObj(c)); 
     }
     else{
       return {[e.id]: e }; 
     }
}

在Javascript中是否可行？

Answer 1

您可以使用递归函数，该函数从给定项目中生成对象而不会改变原始数据。

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SELECT VALUE

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function getObjects(array) {
    var object = {};
    array.forEach(function (item) {
        object[item.id] = Object.assign({}, item, { children: getObjects(item.children) });
    });
    return object;
}

var array = [{ id: 1, desc: 'd1', children: [{ id: 1.1, desc: 'd1.1', children: [] }] },        { id: 2, desc: 'd2', children: [] },        { id: 3, desc: 'd3', children: [] }];

console.log(getObjects(array));

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Answer 2

让我们试试这段代码......

let array = [{id: 1, desc: 'd1', children:[{id:1.1, desc:'d1.1', children:[]}]},
    {id:2, desc:'d2', children:[] },
    {id:3, desc:'d3', children:[] }]

let object = {}

array.forEach(item => {
    let children = item.children
    object[item.id] = item
    object[item.id].children = {}
    children.forEach(child => {
        object[item.id].children[child.id] = child
    })
})

console.log(object)

结果：

{ '1': { id: 1, desc: 'd1', children: { '1.1': [Object] } },
  '2': { id: 2, desc: 'd2', children: {} },
  '3': { id: 3, desc: 'd3', children: {} } }

Answer 3

您可以使用reduce()方法创建递归函数并返回对象对象。

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let array = [{id: 1, desc: 'd1', children:[{id:1.1, desc:'d1.1', children:[]}]}, {id:2, desc:'d2', children:[] }, {id:3, desc:'d3', children:[] }]
                 
function build(data) {
  return data.reduce(function(r, {id, desc, children}) {
    const e = {id, desc}
    if(children && children.length) e.children = build(children);
    r[id] = e
    return r;
  }, {})
}

const result = build(array)
console.log(result)

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