用JSON作为输入创建TypeScript对象最简单的方法是什么?

Question

在Angular2 (TypeScript)中，我有一个具有以下构造函数的类:

 export class DataModel {
    constructor(public date_of_visit: string,
                public gender: boolean,
                public year_of_birth: number,
                public height: number,
                public weight: number){}
 }

我有以下JSON对象:

json = {"date_of_visit": "23/09/2016", "gender": 1, "year_of_birth": 1975, "height":185, "weight": 85}

问题:用JSON数据作为输入创建DataModel实例的最简单方法是什么?比如new DataModel(**json)

Answer 1

对于编译时强制转换，可以这样做:

let dataModel = {"date_of_visit": "23/09/2016", "gender": 1, "year_of_birth": 1975, "height":185, "weight": 85} as DataModel;

Answer 2

选角可以帮你通过:

let model: DataModel = {"date_of_visit": "23/09/2016", "gender": 1, "year_of_birth": 1975, "height":185, "weight": 85} as DataModel;