如何将 JSON 对象解析为 TypeScript 对象

我目前正在尝试将收到的 JSON 对象转换为具有相同属性的 TypeScript 类，但无法使其工作。我究竟做错了什么？

员工阶层

export class Employee{
    firstname: string;
    lastname: string;
    birthdate: Date;
    maxWorkHours: number;
    department: string;
    permissions: string;
    typeOfEmployee: string;
    note: string;
    lastUpdate: Date;
}

员工字符串

{
    "department": "<anystring>",
    "typeOfEmployee": "<anystring>",
    "firstname": "<anystring>",
    "lastname": "<anystring>",
    "birthdate": "<anydate>",
    "maxWorkHours": <anynumber>,
    "username": "<anystring>",
    "permissions": "<anystring>",
    "lastUpdate": "<anydate>"
    //I will add note later
}

我的尝试

let e: Employee = new Employee();

Object.assign(e, {
    "department": "<anystring>",
    "typeOfEmployee": "<anystring>",
    "firstname": "<anystring>",
    "lastname": "<anystring>",
    "birthdate": "<anydate>",
    "maxWorkHours": 3,
    "username": "<anystring>",
    "permissions": "<anystring>",
    "lastUpdate": "<anydate>"
});

console.log(e);

链接到 Typescript Playground https://www.typescriptlang.org/play/#src=class%20Employee%7B%0A%20%20%20%20firstname%3A%20string%3B%0A%20%20%20%20lastname%3A%20string%3B%0A%20%20%20%20birthdate%3A%20Date%3B%0A%20%20%20%20maxWorkHours%3A%20number%3B%0A%20%20%20%20department%3A%20string%3B%0A%20%20%20%20permissions%3A%20string%3B%0A%20%20%20%20typeOfEmployee%3A%20string%3B%0A%20%20%20%20note%3A%20string%3B%0A%20%20%20%20lastUpdate%3A%20Date%3B%0A%7D%0A%0Avar%20e%3A%20Employee%20%3D%20new%20Employee()%3B%0AObject.assign(e%2C%20%7B%0A%20%20%20%20%22department%22%3A%20%22%3Canystring%3E%22%2C%0A%20%20%20%20%22typeOfEmployee%22%3A%20%22%3Canystring%3E%22%2C%0A%20%20%20%20%22firstname%22%3A%20%22%3Canystring%3E%22%2C%0A%20%20%20%20%22lastname%22%3A%20%22%3Canystring%3E%22%2C%0A%20%20%20%20%22birthdate%22%3A%20%22%3Canydate%3E%22%2C%0A%20%20%20%20%22maxWorkHours%22%3A%203%2C%0A%20%20%20%20%22username%22%3A%20%22%3Canystring%3E%22%2C%0A%20%20%20%20%22permissions%22%3A%20%22%3Canystring%3E%22%2C%0A%20%20%20%20%22lastUpdate%22%3A%20%22%3Canydate%3E%22%0A%7D)%3B%0Aconsole.log(e)%3B%0A

如果您使用打字稿界面 https://www.typescriptlang.org/docs/handbook/typescript-in-5-minutes.html#defining-types代替class，事情就更简单了：

export interface Employee {
    typeOfEmployee_id: number;
    department_id: number;
    permissions_id: number;
    maxWorkHours: number;
    employee_id: number;
    firstname: string;
    lastname: string;
    username: string;
    birthdate: Date;
    lastUpdate: Date;
}

let jsonObj = JSON.parse(employeeString); // string to "any" object first
let employee = jsonObj as Employee;

如果你想要一个class但是，简单的转换是行不通的。例如：

class Foo {
    name: string;
    public pump() { }
}

let jsonObj = JSON.parse('{ "name":"hello" }');
let fObj = jsonObj as Foo;
fObj.pump(); // crash, method is undefined!

对于一个类，您必须编写一个接受 JSON 字符串/对象的构造函数，然后迭代属性以手动分配每个成员，如下所示：

class Foo {
    name: string;

    constructor(jsonStr: string) {
        let jsonObj = JSON.parse(jsonStr);
        for (let prop in jsonObj) {
            this[prop] = jsonObj[prop];
        }
    }
}

let fObj = new Foo(theJsonString);