ES6 arrow function and ternary operator:
Object.entries(obj).reduce((acc, entry) => {
   const [key, value] = entry
  if (value !== undefined) acc[key] = value;
  return acc;
}, {})


    const obj = {test:undefined, test1:1 ,test12:0, test123:false};
    const newObj = Object.entries(obj).reduce((acc, entry) => {
       const [key, value] = entry
      if (value !== undefined) acc[key] = value;
      return acc;
    }, {})
    console.log(newObj)

我在我的项目中有同样的场景，并使用以下方法实现。

它适用于所有数据类型，上面提到的一些数据类型不适用于日期和空数组。

removeEmptyKeysFromObject.js

removeEmptyKeysFromObject(obj) { Object.keys(obj).forEach(key => { if (Object.prototype.toString.call(obj[key]) === '[object Date]' && (obj[key].toString().length === 0 || obj[key].toString() === 'Invalid Date')) { delete obj[key]; } else if (obj[key] && typeof obj[key] === 'object') { this.removeEmptyKeysFromObject(obj[key]); } else if (obj[key] == null || obj[key] === '') { delete obj[key]; } if (obj[key] && typeof obj[key] === 'object' && Object.keys(obj[key]).length === 0 && Object.prototype.toString.call(obj[key]) !== '[object Date]') { delete obj[key]; } }); return obj; }

将任何对象传递给该函数

您可能正在寻找delete关键字。

var obj = { };
obj.theProperty = 1;
delete obj.theProperty;

Oneliner:

let obj = { a: 0, b: "string", c: undefined, d: null };

Object.keys(obj).map(k => obj[k] == undefined ? delete obj[k] : obj[k] );

控制台.log（卷）;

Obj将是{a: 0, b: "string"}

你可以使用空合并运算符:??因为它只检查空值和未定义值。注意，下面的例子改变了obj本身。它还删除嵌套对象的空值和未定义值。

const removeEmptyKeys = (obj) => {
    Object.entries(obj).forEach(([k, v]) => {
        (v ?? delete obj[k])
        if (v && typeof v === 'object') {
            removeEmptyKeys(v)
        }
    })
}

使用Nullish合并可用ES2020

const filterNullishPropertiesFromObject = (obj) => {
  const newEntries = Object.entries(obj).filter(([_, value]) => {
    const nullish = value ?? null;
    return nullish !== null;
  });

  return Object.fromEntries(newEntries);
};