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

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    

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)));
}

最后我这样做了:

public clone(): any {
  const result = new (<any>this.constructor);

  // some deserialization code I hade in place already...
  // which deep copies all serialized properties of the
  // object graph
  // result.deserialize(this)

  // you could use any of the usggestions in the other answers to
  // copy over all the desired fields / properties

  return result;
}

因为:

var cloneObj = new (<any>this.constructor());

@Fenton给出了运行时错误。

Typescript版本:2.4.2

我尝试创建一个通用的复制/克隆服务，为嵌套对象保留类型。会爱反馈，如果我做错了什么，但它似乎工作到目前为止…

import { Injectable } from '@angular/core';

@Injectable()
export class CopyService {

  public deepCopy<T>(objectToClone: T): T {
    // If it's a simple type or null, just return it.
    if (typeof objectToClone === 'string' ||
      typeof objectToClone === 'number' ||
      typeof objectToClone === 'undefined' ||
      typeof objectToClone === 'symbol' ||
      typeof objectToClone === 'function' ||
      typeof objectToClone === 'boolean' ||
      objectToClone === null
    ) {
      return objectToClone;
    }

    // Otherwise, check if it has a constructor we can use to properly instantiate it...
    let ctor = Object.getPrototypeOf(objectToClone).constructor;
    if (ctor) {
      let clone = new ctor();

      // Once we've instantiated the correct type, assign the child properties with deep copies of the values
      Object.keys(objectToClone).forEach(key => {
        if (Array.isArray(objectToClone[key]))
          clone[key] = objectToClone[key].map(item => this.deepCopy(item));
        else
          clone[key] = this.deepCopy(objectToClone[key]);
      });

      if (JSON.stringify(objectToClone) !== JSON.stringify(clone))
        console.warn('object cloned, but doesnt match exactly...\nobject: ' + JSON.stringify(objectToClone) + "\nclone: " + JSON.stringify(clone))

      // return our cloned object...
      return clone;
    }
    else {
      //not sure this will ever get hit, but figured I'd have a catch call.
      console.log('deep copy found something it didnt know: ' + JSON.stringify(objectToClone));
      return objectToClone;
    }
  }
}

我在克隆时使用以下方法。它处理我需要的几乎所有东西，甚至将函数复制到新创建的对象。

  public static clone<T>(value: any) : T {
    var o: any = <any>JSON.parse(JSON.stringify(value));
    var functions = (<String[]>Object.getOwnPropertyNames(Object.getPrototypeOf(value))).filter(a => a != 'constructor');
    for (var i = 0; i < functions.length; i++) {
      var name = functions[i].toString();
      o[name] = value[name];
    }
    return <T>o;
  }