我有以下打字稿代码:
class Foo {
private _id: number;
private _desc: string;
constructor(id: number, desc: string) {
this._id = id;
this._desc = desc;
}
public get id(): number {
return this.id;
}
public set id(value: number) {
this._id = value;
}
public get desc():string {
return this.desc;
}
public set desc(value: string) {
this._desc = value;
}
}
let foo = new Foo(1, 'something');
我想从Typescript类中获取一个字符串,我的意思是使用getter和setter。然后我应该得到以下字符串:
{"id":1,"desc":"something"}
根据此answer,我可以达到以下目的:在类中添加以下方法:
public toJSONString(): string {
return JSON.stringify(this, Object.keys(this.constructor.prototype));
}
有效。
如果Typescript类包含任何其他子类,则此方法不起作用。
所以,如果我有以下代码:
class Foo {
private _id: number;
private _desc: string;
private _user: Bar;
constructor(id: number, desc: string, user: Bar) {
this._id = id;
this._desc = desc;
this._user = user;
}
public get id(): number {
return this._id;
}
public set id(value: number) {
this._id = value;
}
public get desc():string {
return this._desc;
}
public set desc(value: string) {
this._desc = value;
}
public get user(): Bar {
return this._user;
}
public set user(value: Bar) {
this._user = value;
}
public toJSONString(): string {
return JSON.stringify(this, Object.keys(this.constructor.prototype));
}
}
class Bar {
private _name: string;
private _surname: string;
constructor(name: string, surname: string) {
this._name = name;
this._surname = surname;
}
public get name(): string {
return this._name;
}
public set name(value: string) {
this._name = value;
}
public get surname():string {
return this._surname;
}
public set surname(value: string) {
this._surname = value;
}
}
let foo = new Foo(1, 'something', new Bar('foo', 'bar'));
如果我使用toJSONString
方法,则会得到以下字符串:
{"id":1,"desc":"something","user":{}}
代替此:
{"id":1,"desc":"something","user":{ "name": "foo", "surname": "bar"}}
那么,如何从具有其他子类的Typescript类中获取字符串?
(如果需要,这里是第一个代码的playground,这里是第二个代码的playground)
答案 0 :(得分:1)
这里有两件事要记住:
定义getter和setter时,它们一旦转换为Javascript便不会成为实例方法,而是使用Object.defineProperty将其添加到原型中。这意味着您不会仅仅使用JSON.stringify
将replacer数组传递给JSON.stringify并告诉它仅使用原型值可以完成此工作,但它对嵌套对象不起作用。事实是JSON.stringify将仅解析具有该名称的属性,无论它位于对象结构中的什么位置。
例如
let a = {
user: "Foo",
data: {
name: "Bar"
}
};
JSON.stringify(a, ["user", "data"]);
将输出{"user":"Foo","data":{}}
,因为即使嵌套对象的键是data
,对象本身也不具有名为user
或data
的属性
但是
let a = {
user: "Foo",
data: {
user: "Bar"
}
};
JSON.stringify(a, ["user", "data"]);
将输出{"user":"Foo","data":{"user":"Bar"}}
,因为嵌套对象具有一个称为user
的属性,就像它的父对象一样
我认为这种行为可能会造成混乱,但是可以通过创建一种方法来实现解决方案,该方法将获取您感兴趣的所有对象的所有属性。我没有在Typescript中找到检查类是否实现的方法一个接口(或扩展一个类),所以即使它不是那么“优雅”,我也必须对我所知道的东西进行一些工作。
abstract class Stringifyiable {
private isStringifyiable(value): boolean {
return value != null && (typeof value === 'object' || typeof value === 'function') && value['getJsonKeys'] && typeof value['getJsonKeys'] === 'function';
}
public getJsonKeys(): string[] {
let keys = Object.keys(this.constructor.prototype);
keys.forEach(key => {
if (this.isStringifyiable(this[key])) {
keys = keys.concat(this[key].getJsonKeys());
}
});
return keys;
}
public toJSONString(): string {
return JSON.stringify(this, this.getJsonKeys());
}
}
class Foo extends Stringifyiable {
private _id: number;
private _desc: string;
private _user: Bar;
constructor(id: number, desc: string, user: Bar) {
super();
this._id = id;
this._desc = desc;
this._user = user;
}
public get id(): number {
return this._id;
}
public set id(value: number) {
this._id = value;
}
public get desc():string {
return this._desc;
}
public set desc(value: string) {
this._desc = value;
}
public get user(): Bar {
return this._user;
}
public set user(value: Bar) {
this._user = value;
}
}
class Bar extends Stringifyiable {
private _name: string;
private _surname: string;
constructor(name: string, surname: string) {
super();
this._name = name;
this._surname = surname;
}
public get name(): string {
return this._name;
}
public set name(value: string) {
this._name = value;
}
public get surname():string {
return this._surname;
}
public set surname(value: string) {
this._surname = value;
}
}
let foo = new Foo(1, 'something', new Bar('foo', 'bar'));
//this will output {"id":1,"desc":"something","user":{"name":"foo","surname":"bar"}}
foo.toJSONString();
在使用循环引用时要小心,因为它会陷入无限循环(不过我敢肯定它可以解决)。