Question

我想获取一个对象数组并过滤每个对象，以仅返回键与数组中的某个项目匹配的属性。

例如：

const myKeys    = ['key_1', 'key_3'];
const myArray   = [
    { 
        key_1: 'Some Value A',
        key_2: 'Some Other Value A', 
        key_3: 'Some Final Value A',
    },
    { 
        key_1: 'Some Value B',
        key_2: 'Some Other Value B', 
        key_3: 'Some Final Value B',
    },
    { 
        key_1: 'Some Value C',
        key_2: 'Some Other Value C', 
        key_3: 'Some Final Value C',
    },
];

应产生以下结果：

const result   = [
    { 
        key_1: 'Some Value A',
        key_3: 'Some Final Value A',
    },
    { 
        key_1: 'Some Value B',
        key_3: 'Some Final Value B',
    },
    { 
        key_1: 'Some Value C',
        key_3: 'Some Final Value C',
    },
];

我相信结合使用.map()和.filter()可以做到这一点，但是我不确定如何最好地使用ES6达到以下效果：

const filteredData = (array, keys) => {
    const newArr = [];

    for (let i = 0; i < myArray.length; i++) {
        const item      = myArray[i];
        let newObj      = {};

        for (let j = 0; j < myKeys.length; j++) {
            const filter = myKeys[j];
            newObj[filter] = item[filter];
        }

        newArr.push(newObj);
    }

    return newArr;
}

console.log(filteredData(myArray, myKeys));

Answer 1

您可以映射所拥有的条目并使用它来构建对象。

const 
    keys = ['key_1', 'key_3'],
    data = [{ key_1: 'Some Value A', key_2: 'Some Other Value A', key_3: 'Some Final Value A' }, { key_1: 'Some Value B', key_2: 'Some Other Value B', key_3: 'Some Final Value B' }, { key_1: 'Some Value C', key_2: 'Some Other Value C', key_3: 'Some Final Value C' }],
    result = data.map(o => Object.fromEntries(keys.map(k => [k, o[k]])));

console.log(result);

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Answer 2

您可以简单地遍历keys并构建一个仅包含这些键和相应值的新对象

const myKeys = ['key_1', 'key_3'];
const myArray = [{key_1: 'Some Value A',key_2: 'Some Other Value A',key_3: 'Some Final Value A',},{key_1: 'Some Value B',key_2: 'Some Other Value B',key_3: 'Some Final Value B',},{key_1: 'Some Value C',key_2: 'Some Other Value C',key_3: 'Some Final Value C',},];

let final = myArray.map(v => {
  return myKeys.reduce((op, key) => {
    op[key] = v[key]
    return op
  }, {})
})

console.log(final)

Answer 3

您可以将每个对象.map移至其条目，以便可以.filter移出任何myKeys中没有键的条目，然后使用.fromEntries()进行构建像这样设置新对象：

const myKeys = ['key_1', 'key_3'];
const myArray = [{
    key_1: 'Some Value A',
    key_2: 'Some Other Value A',
    key_3: 'Some Final Value A',
  },
  {
    key_1: 'Some Value B',
    key_2: 'Some Other Value B',
    key_3: 'Some Final Value B',
  },
  {
    key_1: 'Some Value C',
    key_2: 'Some Other Value C',
    key_3: 'Some Final Value C',
  },
];

const res = myArray.map(obj => Object.fromEntries(Object.entries(obj).filter(([k, v]) => myKeys.includes(k))));

console.log(res);

JS ES6通过键数组过滤对象的正确方法

4 个答案: