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Javascript Typescript:使用fromJson方法指定的对象,如何测试私有属性是否存在并设置它_Javascript_Typescript_Ecmascript 6 - Fatal编程技术网

Javascript Typescript:使用fromJson方法指定的对象,如何测试私有属性是否存在并设置它

Javascript Typescript:使用fromJson方法指定的对象,如何测试私有属性是否存在并设置它,javascript,typescript,ecmascript-6,Javascript,Typescript,Ecmascript 6,我有一个包含fromJson方法的对象。由于无法访问类的私有属性,此方法不起作用?出了什么问题以及如何处理?代码是用TypeScript编写的 class Example { private Foo: string; // does not matter if private or public, same effect, and normaly has to be private constructor(input?: string) { if (!!inpu

我有一个包含fromJson方法的对象。由于无法访问类的私有属性,此方法不起作用?出了什么问题以及如何处理?代码是用TypeScript编写的

class Example {
    private Foo: string; // does not matter if private or public, same effect, and normaly has to be private

    constructor(input?: string) {
        if (!!input) {
            this.foo = input;    
        }
    }

    set foo(value: string) {
        this.Foo = value;
    }
    get foo(): string {
        return this.Foo;
    }

    public static fromJson(obj: Object) {
        let result: Example = new Example();
        for (let index in obj) {
            if (Example.hasOwnProperty(index)) { // never runs because false
                result[index] = obj[index];
            }

            /* allready tried this -> same result */
            // if (result.hasOwnProperty(index)) {
            //    result[index] = obj[index];
            //}

            // let descriptor = Object.getOwnPropertyDescriptor(Example, index); // = undefined
            // let descriptor = Object.getOwnPropertyDescriptor(result, index); // = undefined
        }
        return result;
    }

    public toJsonString() {
        return JSON.stringify(this);
    }

    public toJsonObject() {
        return JSON.parse(this.toJsonString());
    }
}

let a = new Example('one');
let json = a.toJsonObject();  // this looks exactly like my api response (type json)
let obj = Example.fromJson(json);
console.log(json);
console.log(obj);
但是
console.log(obj)
必须是
{“Foo”:“one”,Foo(…)}

编辑:生成的JavaScript:

var Example = (function () {
    function Example(input) {
        if (!!input) {
            this.foo = input;
        }
    }
    Object.defineProperty(Example.prototype, "foo", {
        get: function () {
            return this.Foo;
        },
        set: function (value) {
            this.Foo = value;
        },
        enumerable: true,
        configurable: true
    });
    Example.fromJson = function (obj) {
        var result = new Example();
        for (var index in obj) {
            if (Example.hasOwnProperty(index)) {
                result[index] = obj[index];
            }
        }
        return result;
    };
    Example.prototype.toJsonString = function () {
        return JSON.stringify(this);
    };
    Example.prototype.toJsonObject = function () {
        return JSON.parse(this.toJsonString());
    };
    return Example;
}());
var a = new Example('one');
var json = a.toJsonObject(); // this looks exactly like my api response (type json)
var obj = Example.fromJson(json);
console.log(json);
console.log(obj);

我认为这就是你正在寻找的解决方案。如果您用未定义的属性初始化属性,那么toJson方法不会列出参数。因此,您的请求流量没有那么大。

生成的javascript看起来像什么?@Arg0n通过edit质疑如果将值分配给
Foo
,生成的JS会发生什么<代码>私有Foo:string=null氩气是正确的。必须初始化每个属性才能通过hasOwnProperty函数访问。我发布了一个工作样本,这正是我想要的。现在初始化我所有模型中的所有属性是一件愚蠢的手工工作;)
class Example {
    private Foo: string = undefined;
    private Foo2: number = undefined;

    constructor(input?: string) {
        if (!!input) {
            this.foo = input;    
        }
    }

    set foo(value: string) {
        this.Foo = value;
    }
    get foo(): string {
        return this.Foo;
    }

    set numeric(value: number) {
        this.Foo2 = value;
    }
    get numeric(): number {
        return this.Foo2;
    }

    public static fromJson(obj: Object) {
        let result: Example = new Example();
        for (let index in obj) {
            if (result.hasOwnProperty(index)) {
                result[index] = obj[index]; // care, has to be result
            }
        }
        return result;
    }

    public toJsonString() {
        return JSON.stringify(this);
    }

    public toJsonObject() {
        return JSON.parse(this.toJsonString());
    }
}

let a = new Example('one');
let json = a.toJsonObject();
let obj = Example.fromJson(json);
console.log(json);
console.log(obj);