如何合并两个javascript对象的属性,而不是使用空值?

function merge(obj1, obj2) {
  answer = {}
  for(key in obj1) {
    if(answer[key] === undefined || answer[key] === null)
      answer[key] = obj1[key];
  }
  for(key in obj2) {
    if(answer[key] === undefined || answer[key] === null)
      answer[key] = obj2[key];
  }
  return answer
}

function omitNull (obj) {
  const keys = Object.keys(obj);
  const _obj = {};
  for (const key of keys) {
    const value = obj[key];
    if (value !== null) {
      _obj[key] = value;
    }
  }
  return _obj;
}

const merged = {
  ...omitNull(obj1),
  ...omitNull(obj2)
};

result = {};
for(key in obj1){
    result[key] = obj1[key] === null ? obj2[key] : obj1[key];
}

这将两个对象合并在一个循环中

试试这个。

const obj1 = {a: null, b: "b"} 
const obj2 = {a: "a", b: null}
const obj3 = {}
for (var k in obj1) {
  obj3[k] = obj1[k] ? obj1[k] : obj2[k];
}
console.log(obj3);

简单的怎么样 forEach() 对象上的循环 key . 两个都能用 null 和 undefined 值:

const obj1 = {a: null, b: "b"};
const obj2 = {a: "a", b: null};
const merged = {};
Object.keys(obj1).forEach((key) => merged[[key]] = obj1[key] ? obj1[key] : obj2[key]);
console.log(merged);

您可以创建一个函数,该函数将对象数组作为参数。

这样,无论您拥有多少对象,您都将得到合并在一起的结果,不包括未定义和空值。只需将它们作为数组发送。

在这里,您传递所有对象,映射它们,然后迭代它们的键和值 for (const [key, value] of Object.entries(obj)) 排除那些 undefined 或 null

见下文

const obj1 = {
  a: null,
  b: "goodb",
  c: 0,
}
const obj2 = {
  a: "gooda",
  b: null,
  c: undefined
}

function cleanObjects(arr) {
  let o = {}
  arr.map((obj) => {
    for (const [key, value] of Object.entries(obj)) {
      typeof value === 'undefined' || value === null ?
      	delete obj[key] : o[key] = value;
    }
  })
  return o;
}
const result = cleanObjects([obj1, obj2])
console.log(result)