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
自从TypeScript 3.7发布以来，现在支持递归类型别名，它允许我们定义一个类型安全的deepCopy()函数:
// DeepCopy type can be easily extended by other types,
// like Set & Map if the implementation supports them.
type DeepCopy<T> =
T extends undefined | null | boolean | string | number ? T :
T extends Function | Set<any> | Map<any, any> ? unknown :
T extends ReadonlyArray<infer U> ? Array<DeepCopy<U>> :
{ [K in keyof T]: DeepCopy<T[K]> };
function deepCopy<T>(obj: T): DeepCopy<T> {
// implementation doesn't matter, just use the simplest
return JSON.parse(JSON.stringify(obj));
}
interface User {
name: string,
achievements: readonly string[],
extras?: {
city: string;
}
}
type UncopiableUser = User & {
delete: () => void
};
declare const user: User;
const userCopy: User = deepCopy(user); // no errors
declare const uncopiableUser: UncopiableUser;
const uncopiableUserCopy: UncopiableUser = deepCopy(uncopiableUser); // compile time error
操场上
自从TypeScript 3.7发布以来，现在支持递归类型别名，它允许我们定义一个类型安全的deepCopy()函数:
// DeepCopy type can be easily extended by other types,
// like Set & Map if the implementation supports them.
type DeepCopy<T> =
T extends undefined | null | boolean | string | number ? T :
T extends Function | Set<any> | Map<any, any> ? unknown :
T extends ReadonlyArray<infer U> ? Array<DeepCopy<U>> :
{ [K in keyof T]: DeepCopy<T[K]> };
function deepCopy<T>(obj: T): DeepCopy<T> {
// implementation doesn't matter, just use the simplest
return JSON.parse(JSON.stringify(obj));
}
interface User {
name: string,
achievements: readonly string[],
extras?: {
city: string;
}
}
type UncopiableUser = User & {
delete: () => void
};
declare const user: User;
const userCopy: User = deepCopy(user); // no errors
declare const uncopiableUser: UncopiableUser;
const uncopiableUserCopy: UncopiableUser = deepCopy(uncopiableUser); // compile time error
操场上
通过在TypeScript 2.1中引入的“Object Spread”，很容易获得一个浅拷贝
this TypeScript: let copy ={…原始};
生成这个JavaScript:
var __assign = (this && this.__assign) || Object.assign || function(t) {
for (var s, i = 1, n = arguments.length; i < n; i++) {
s = arguments[i];
for (var p in s) if (Object.prototype.hasOwnProperty.call(s, p))
t[p] = s[p];
}
return t;
};
var copy = __assign({}, original);
https://www.typescriptlang.org/docs/handbook/release-notes/typescript-2-1.html
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
}
