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
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}
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
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)
我自己也遇到过这个问题,最后写了一个小库clone -ts,它提供了一个抽象类,它向任何扩展它的类添加了一个克隆方法。抽象类借用了芬顿接受的答案中描述的深度复制函数,只是替换了Copy = {};使用copy = object. create(originalObj)来保留原始对象的类。下面是一个使用该类的示例。
import {Cloneable, CloneableArgs} from 'cloneable-ts';
// Interface that will be used as named arguments to initialize and clone an object
interface PersonArgs {
readonly name: string;
readonly age: number;
}
// Cloneable abstract class initializes the object with super method and adds the clone method
// CloneableArgs interface ensures that all properties defined in the argument interface are defined in class
class Person extends Cloneable<TestArgs> implements CloneableArgs<PersonArgs> {
readonly name: string;
readonly age: number;
constructor(args: TestArgs) {
super(args);
}
}
const a = new Person({name: 'Alice', age: 28});
const b = a.clone({name: 'Bob'})
a.name // Alice
b.name // Bob
b.age // 28
或者你可以直接用克隆。克隆助手方法:
import {Cloneable} from 'cloneable-ts';
interface Person {
readonly name: string;
readonly age: number;
}
const a: Person = {name: 'Alice', age: 28};
const b = Cloneable.clone(a, {name: 'Bob'})
a.name // Alice
b.name // Bob
b.age // 28