从具有相同键的两个 Javascript 个对象中提取一个公共对象

Question

我有两个具有相同键的 Javascript 对象，但 obj1 中具有某些值的键在 obj2 中为空反之亦然。两个对象中也可以有空键。

obj1 = {
 a: 1,
 b: 2,
 c: ' ',
 d: ' ',
 e: ' '
}

obj2 = {
 a: ' ',
 b: ' ',
 c: 3,
 d: 5,
 e: ' '
}

我想要一个组合对象，其中包含具有来自两个对象的值的键，如果两个键都是空的，它只是将其放置为空

Expected result:

result = {
 a: 1,
 b: 2,
 c: 3,
 d: 5,
 e: ' '
}

我试过使用扩展运算符，但第二个对象总是优先于第一个对象：> {...obj1, ...obj2}

Answer 1

您必须构建一个通用的组合对象，然后遍历每个其他对象，如果不为空则添加值

obj1 = {
    a: 1,
    b: 2,
    c: ' ',
    d: ' ',
    e: ' '
}

obj2 = {
    a: ' ',
    b: ' ',
    c: 3,
    d: 5,
    e: ' '
}

// Combine the two to get all keys into the combined object
var combined = { ...obj1, ...obj2 };

// Set each value in the combined object to a single spaced string
for (let key in combined) {
    combined[key] = ' ';
}

// Loop through each item in obj1 and add to combined if a single spaced string
for (let key in obj1) {
    if (obj1[key] !== ' ') combined[key] = obj1[key];
}

// Loop through each item in obj2 and add to combined if a single spaced string
for (let key in obj2) {
    if (obj2[key] !== ' ') combined[key] = obj2[key];
}

console.log(combined)

Answer 2

您可以利用 Object.keys 方法和 reduce 方法来创建结果对象

let obj1 = {
  a: 1,
  b: 2,
  c: ' ',
  d: ' ',
  e: ' '
}

let obj2 = {
  a: ' ',
  b: ' ',
  c: 3,
  d: 5,
  e: ' '
}
let result = Object.keys(obj1).reduce((acc, curr) => {
  val = obj1[curr] != " " ? obj1[curr] : obj2[curr];
  return { ...acc,
    ...{
      [curr]: val
    }
  }
}, {})

console.log(result)

Answer 3

Object.fromEntries(Object.entries(obj1).map(e=>([e[0], obj1[e[0]]!=' '?obj1[e[0]]:obj2[e[0]]])))

Answer 4

您可以过滤对象并传播第一个和过滤后的对象。

const
    filter = o => Object.fromEntries(Object.entries(o).filter(([, v]) => v !== ' ')),
    obj1 = { a: 1, b: 2, c: ' ', d: ' ', e: ' ' },
    obj2 = { a: ' ', b: ' ', c: 3, d: 5, e: ' ' },
    result = { ...obj1, ...filter(obj2) };

console.log(result);

Answer 5

var obj1 = {
    a: 1,
    b: 2,
    c: ' ',
    d: ' ',
    e: ' '
}

var obj2 = {
    a: ' ',
    b: ' ',
    c: 3,
    d: 5,
    e: ' '
}

/* in case your objects are not same length */
var longObj = obj1;
var shortObj = obj2;

if(Object.keys(obj1).length < Object.keys(obj2).length) {
  longObj = obj2;
  shortObj = obj1;
}

for(let k in shortObj) {
  longObj[k] = longObj[k] !== ' ' ? longObj[k] : shortObj[k]; 
}

console.log(longObj);

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