Javascript 如何让 .map() 返回新数组及其初始文本属性而不是编号索引？

Question

我正在尝试展平对象的嵌套数组。

我有以下数组:

array = [
  {
    id: 123,
    name: 'Peter',
    phone: '',
    addresses:
      [
        {
          address1: 'Manchester, UK', address2: 'London, UK'
        },
        {
          address1: 'Liverpool, UK', address2: 'NY, USA'
        }
      ]
  },
  {
    id: 124,
    name: 'Sara',
    phone: '',
    addresses: [{ address1: 'London, UK', address2: 'Paris, FR' }]
  }
];

所需的输出应该是这样的:

array = [
  {
    id: 123,
    name: 'Peter',
    phone: '',
    addresses_address1: 'Manchester, UK',
    addresses_address2: 'London, UK'
  },
  {
    id: 124,
    name: 'Sara',
    phone: '',
    addresses_address1: 'London, UK',
    addresses_address2: 'Paris, FR'
  }
];

我使用了.map和.reduce函数，示例取自this site :

var flatten = function (data) {
  return data.reduce(function iter(r, a) {
    if (typeof a === 'object') {
      return Object.keys(a).map(k => a[k]).reduce(iter, r);
    }
    return r.concat(a)
  }, []);
}
console.log(flatten(this.array))

但是它不是获取带有文本属性 id name... 的数组，而是返回索引 0, 1 ...

这是一个stackblitz .

Answer 1

您可以将索引作为属性的一部分，并为外部数组的每个元素返回一个新对象。

function flat(object, path = []) {
    return object && typeof object === 'object'
        ? Object
            .entries(object)
            .reduce((r, [k, v]) => ({ ...r, ...flat(v, [...path, k]) }), {})
        : { [path.join('-')]: object };
}

var array = [{ id: 123, name: 'Peter', phone: '', addresses: [{ address1: 'Manchester, UK', address2: 'London, UK' }, { address1: 'Liverpool, UK', address2: 'NY, USA' }] }, { id: 124, name: 'Sara', phone: '', addresses: [{ address1: 'London, UK', address2: 'Paris, FR' }] }],
    result = array.map(o => flat(o));

console.log(result);