如何使用es6展平嵌套的对象数组

Question

我有这个对象数组，其中我有另一个对象数组，如何获取：

[
  { id: "5a60626f1d41c80c8d3f8a85" },
  { id: "5a6062661d41c80c8b2f0413" },
  { id: "5a60626f1d41c80c8d3f8a83" },
  { id: "5a60626f1d41c80c8d3f8a84" }
];

来自：

[
  {
    id: 1,
    country: [
      {
        id: "5a60626f1d41c80c8d3f8a85"
      },
      {
        id: "5a6062661d41c80c8b2f0413"
      }
    ]
  },
  {
    id: 2,
    country: [
      {
        id: "5a60626f1d41c80c8d3f8a83"
      },
      {
        id: "5a60626f1d41c80c8d3f8a84"
      }
    ]
  }
];

不使用forEach和临时变量？

当我这样做时：

(data || []).map(o=>{
  return o.country.map(o2=>({id: o2.id}))
})

我得到了相同的结构。

Answer 1

最新编辑

所有现代 JS 环境现在都支持Array.prototype.flat和Array.prototype.flatMap

 const data=[{id:1,country:[{id:"5a60626f1d41c80c8d3f8a85"},{id:"5a6062661d41c80c8b2f0413"}]},{id:2,country:[{id:"5a60626f1d41c80c8d3f8a83"},{id:"5a60626f1d41c80c8d3f8a84"}]}]; console.log( data.flatMap( (elem) => elem.country ) )

---

旧答案

不需要任何 ES6 魔法，你可以通过连接内部country数组来减少数组。

 const data=[{id:1,country:[{id:"5a60626f1d41c80c8d3f8a85"},{id:"5a6062661d41c80c8b2f0413"}]},{id:2,country:[{id:"5a60626f1d41c80c8d3f8a83"},{id:"5a60626f1d41c80c8d3f8a84"}]}]; console.log( data.reduce( (arr, elem) => arr.concat(elem.country), [] ) )

如果您想要 ES6 功能（箭头函数除外），请使用数组展开而不是 concat 方法：

 const data=[{id:1,country:[{id:"5a60626f1d41c80c8d3f8a85"},{id:"5a6062661d41c80c8b2f0413"}]},{id:2,country:[{id:"5a60626f1d41c80c8d3f8a83"},{id:"5a60626f1d41c80c8d3f8a84"}]}]; console.log( data.reduce( (arr, elem) => [...arr, ...elem.country], [] ) )

注意：这些建议会在每次迭代时创建一个新数组。

为了效率，你必须牺牲一些优雅：

 const data=[{id:1,country:[{id:"5a60626f1d41c80c8d3f8a85"},{id:"5a6062661d41c80c8b2f0413"}]},{id:2,country:[{id:"5a60626f1d41c80c8d3f8a83"},{id:"5a60626f1d41c80c8d3f8a84"}]}]; console.log( data.reduce( (arr, elem) => { for (const c of elem.country) { arr.push(c); } return arr; }, [] ) )

Answer 2

 const raw = [ { id: 1, country: [ { id: "5a60626f1d41c80c8d3f8a85" }, { id: "5a6062661d41c80c8b2f0413" } ] }, { id: 2, country: [ { id: "5a60626f1d41c80c8d3f8a83" }, { id: "5a60626f1d41c80c8d3f8a84" } ] } ]; const countryIds = raw .map(x => x.country) .reduce((acc, curr) => { return [ ...acc, ...curr.map(x => x.id) ]; }, []); console.log(countryIds)

Answer 3

这行得通，只需连接您的解决方案返回的嵌套数组

 let arr = [{ "id": 1, "country": [{ "id": "5a60626f1d41c80c8d3f8a85", }, { "id": "5a6062661d41c80c8b2f0413", } ] }, { "id": 2, "country": [{ "id": "5a60626f1d41c80c8d3f8a83", }, { "id": "5a60626f1d41c80c8d3f8a84", } ] } ]; //If you want an array of country objects console.log([].concat.apply(...(arr || []).map(o=> o.country))) //If you can an array od country ids console.log([].concat.apply(...(arr || []).map(o=> o.country.map(country => country.id))))

Answer 4

Ayush Gupta 的解决方案适用于这种情况。 但我想提供其他解决方案。

 const arr = [ { id: 1, country: [ { id: '5a60626f1d41c80c8d3f8a85' }, { id: '5a6062661d41c80c8b2f0413' } ] }, { id: 2, country: [ { id: '5a60626f1d41c80c8d3f8a83' }, { id: '5a60626f1d41c80c8d3f8a84' } ] } ]; const ids = arr.reduce( (acc, {country}) => [ ...acc, ...country.map(({id}) => ({ id })) ], [] ); console.log(ids);

Answer 5

对于 JSON 字符串数据，也可以在解析过程中完成：

 var ids = [], json = '[{"id":1,"country":[{"id":"5a60626f1d41c80c8d3f8a85"},{"id":"5a6062661d41c80c8b2f0413"}]},{"id":2,"country":[{"id":"5a60626f1d41c80c8d3f8a83"},{"id":"5a60626f1d41c80c8d3f8a84"}]}]'; JSON.parse(json, (k, v) => v.big && ids.push(v)); console.log( ids );

Answer 6

我不知道为什么没有人提到flat() （可能对于大型阵列，它的性能可能较低）

(data || []).map(o=>{
  return o.country.map(o2=>({id: o2.id}))
}).flat()