Typescript克隆对象

我有一个超类，它是许多子类(Customer, Product, ProductCategory…)的父类(Entity)。

我想在Typescript中动态克隆一个包含不同子对象的对象。

例如:拥有不同产品的客户拥有一个ProductCategory

var cust:Customer  = new Customer ();

cust.name = "someName";
cust.products.push(new Product(someId1));
cust.products.push(new Product(someId2));

为了克隆对象的整个树，我在实体中创建了一个函数

public clone():any {
    var cloneObj = new this.constructor();
    for (var attribut in this) {
        if(typeof this[attribut] === "object"){
           cloneObj[attribut] = this.clone();
        } else {
           cloneObj[attribut] = this[attribut];
        }
    }
    return cloneObj;
}

当new被转译为javascript时，将引发以下错误:错误TS2351:不能对缺少调用或构造签名的表达式使用'new'。

虽然脚本工作，但我想摆脱转译错误

当前回答

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)

2019-11-18 13:28:33

其他回答

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)

2019-11-18 13:28:33

对于一个简单的克隆的hole对象的内容，我简单地stringify和解析实例:

let cloneObject = JSON.parse(JSON.stringify(objectToClone))

当我在objectToClone树中更改数据时，在cloneObject中没有更改。这就是我的要求。

希望能有所帮助

2017-10-22 09:01:36

function instantiateEmptyObject(obj: object): object {
    if (obj == null) { return {}; }

    const prototype = Object.getPrototypeOf(obj);
    if (!prototype) {
        return {};
    }

    return Object.create(prototype);
}

function quickCopy(src: object, dest: object): object {
    if (dest == null) { return dest; }

    return { ...src, ...dest };
}

quickCopy(src, instantiateEmptyObject(new Customer()));

2020-08-10 16:30:01

下面是deepCopy在TypeScript中的实现(代码中不包含任何内容):

const deepCopy = <T, U = T extends Array<infer V> ? V : never>(source: T ): T => {
  if (Array.isArray(source)) {
    return source.map(item => (deepCopy(item))) as T & U[]
  }
  if (source instanceof Date) {
    return new Date(source.getTime()) as T & Date
  }
  if (source && typeof source === 'object') {
    return (Object.getOwnPropertyNames(source) as (keyof T)[]).reduce<T>((o, prop) => {
      Object.defineProperty(o, prop, Object.getOwnPropertyDescriptor(source, prop)!)
      o[prop] = deepCopy(source[prop])
      return o
    }, Object.create(Object.getPrototypeOf(source)))
  }
  return source
}

2022-08-22 11:51:40

你也可以有这样的东西:

class Entity {
    id: number;

    constructor(id: number) {
        this.id = id;
    }

    clone(): this {
        return new (this.constructor as typeof Entity)(this.id) as this;
    }
}

class Customer extends Entity {
    name: string;

    constructor(id: number, name: string) {
        super(id);
        this.name = name;
    }

    clone(): this {
        return new (this.constructor as typeof Customer)(this.id, this.name) as this;
    }
}

只是要确保在所有Entity子类中覆盖clone方法，否则最终会得到部分克隆。

它的返回类型将始终与实例的类型匹配。

2016-08-02 04:44:17

Typescript克隆对象

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