也许是这样的:

var copy = Object.assign({}, {a: 1, b: 2, c: 3})
delete copy.c;

这样够好吗?或者说c不能被复制?

我用的是ESNext one liner

Const obj = {a: 1, b: 2, c: 3, d: 4} Const clone = (({b, c，…O}) => O)(obj) //删除b和c console.log(克隆)

也许是这样的:

var copy = Object.assign({}, {a: 1, b: 2, c: 3})
delete copy.c;

这样够好吗?或者说c不能被复制?

使用lodash清洁和快速，有了这个解决方案，您可以删除多个键，也不改变原始对象。这在不同的情况下具有更强的可扩展性和可用性:

import * as _ from 'lodash';

function cloneAndRemove(
    removeTheseKeys: string[],
    cloneObject: any,
): object | never {
    const temp = _.cloneDeep(cloneObject);
    removeTheseKeys.forEach((key) => {
        delete temp[key];
    });

    return temp;
}

export { cloneAndRemove };

var clone = Object.assign({}, {a: 1, b: 2, c: 3});
delete clone.b;

或者如果你接受属性为未定义:

var clone = Object.assign({}, {a: 1, b: 2, c: 3}, {b: undefined});

您可以为它编写一个简单的helper函数。Lodash有一个同名的类似函数:省略

function omit(obj, omitKey) {
  return Object.keys(obj).reduce((result, key) => {
    if(key !== omitKey) {
       result[key] = obj[key];
    }
    return result;
  }, {});
}

omit({a: 1, b: 2, c: 3}, 'c')  // {a: 1, b: 2}

另外，注意它比Object快。然后分配和删除:http://jsperf.com/omit-key