对于深度克隆对象，可以包含另一个对象，数组等，我使用:

const clone = <T>(source: T): T => {
  if (source === null) return source

  if (source instanceof Date) return new Date(source.getTime()) as any

  if (source instanceof Array) return source.map((item: any) => clone<any>(item)) as any

  if (typeof source === 'object' && source !== {}) {
    const clonnedObj = { ...(source as { [key: string]: any }) } as { [key: string]: any }
    Object.keys(clonnedObj).forEach(prop => {
      clonnedObj[prop] = clone<any>(clonnedObj[prop])
    })

    return clonnedObj as T
  }

  return source
}

Use:

const obj = {a: [1,2], b: 's', c: () => { return 'h'; }, d: null, e: {a:['x'] }}
const objClone = clone(obj)

对于可序列化的深度克隆，类型信息为，

export function clone<T>(a: T): T {
  return JSON.parse(JSON.stringify(a));
}

你可以在扩展语法中使用解构赋值:

var obj = {id = 1, name = 'product1'};
var clonedObject = {...obj};

下面是一个现代的实现，它也解释了Set和Map:

export function deepClone<T extends object>(value: T): T {
  if (typeof value !== 'object' || value === null) {
    return value;
  }

  if (value instanceof Set) {
    return new Set(Array.from(value, deepClone)) as T;
  }

  if (value instanceof Map) {
    return new Map(Array.from(value, ([k, v]) => [k, deepClone(v)])) as T;
  }

  if (value instanceof Date) {
    return new Date(value) as T;
  }

  if (value instanceof RegExp) {
    return new RegExp(value.source, value.flags) as T;
  }

  return Object.keys(value).reduce((acc, key) => {
    return Object.assign(acc, { [key]: deepClone(value[key]) });
  }, (Array.isArray(value) ? [] : {}) as T);
}

尝试一下:

deepClone({
  test1: { '1': 1, '2': {}, '3': [1, 2, 3] },
  test2: [1, 2, 3],
  test3: new Set([1, 2, [1, 2, 3]]),
  test4: new Map([['1', 1], ['2', 2], ['3', 3]])
});

test1:
  1: 1
  2: {}
  3: [1, 2, 3]

test2: Array(3)
  0: 1
  1: 2
  2: 3

test3: Set(3)
  0: 1
  1: 2
  2: [1, 2, 3]

test4: Map(3)
  0: {"1" => 1}
  1: {"2" => 2}
  2: {"3" => 3}

试试这个:

let copy = (JSON.parse(JSON.stringify(objectToCopy)));

这是一个很好的解决方案，直到您使用非常大的对象或对象具有不可序列化的属性。

为了保持类型安全，你可以在你想复制的类中使用一个copy函数:

getCopy(): YourClassName{
    return (JSON.parse(JSON.stringify(this)));
}

或以静态的方式:

static createCopy(objectToCopy: YourClassName): YourClassName{
    return (JSON.parse(JSON.stringify(objectToCopy)));
}

在typeScript中，我用angular进行了测试，结果还不错

deepCopy(obj) {


        var copy;

        // Handle the 3 simple types, and null or undefined
        if (null == obj || "object" != typeof obj) return obj;

        // Handle Date
        if (obj instanceof Date) {
            copy = new Date();
            copy.setTime(obj.getTime());
            return copy;
        }

        // Handle Array
        if (obj instanceof Array) {
            copy = [];
            for (var i = 0, len = obj.length; i < len; i++) {
                copy[i] = this.deepCopy(obj[i]);
            }
            return copy;
        }

        // Handle Object
        if (obj instanceof Object) {
            copy = {};
            for (var attr in obj) {
                if (obj.hasOwnProperty(attr)) copy[attr] = this.deepCopy(obj[attr]);
            }
            return copy;
        }

        throw new Error("Unable to copy obj! Its type isn't supported.");
    }