如何将JSON对象解析为TypeScript对象

Question

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，但无法使其正常工作。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);

链接到打字稿游乐场

Answer 1

编译器允许您将返回的对象强制JSON.parse转换为类的原因是因为typescript基于结构子类型。
您实际上并没有的实例Employee，而是拥有一个具有相同属性的对象（如在控制台中看到的）。

一个简单的例子：

class A {
    constructor(public str: string, public num: number) {}
}

function logA(a: A) {
    console.log(`A instance with str: "${ a.str }" and num: ${ a.num }`);
}

let a1 = { str: "string", num: 0, boo: true };
let a2 = new A("stirng", 0);
logA(a1); // no errors
logA(a2);

（操场上的代码）

没有错误是因为a1满足类型，A因为它具有所有属性，并且只要它具有相同的属性，logA即使函数接收的实例不是该函数的实例，也可以在没有运行时错误的情况下调用该函数A。

当您的类是简单的数据对象并且没有方法时，这很好用，但是一旦您引入了方法，事情就会崩溃：

class A {
    constructor(public str: string, public num: number) { }

    multiplyBy(x: number): number {
        return this.num * x;
    }
}

// this won't compile:
let a1 = { str: "string", num: 0, boo: true } as A; // Error: Type '{ str: string; num: number; boo: boolean; }' cannot be converted to type 'A'

// but this will:
let a2 = { str: "string", num: 0 } as A;

// and then you get a runtime error:
a2.multiplyBy(4); // Error: Uncaught TypeError: a2.multiplyBy is not a function

（操场上的代码）

编辑

这很好用：

const employeeString = '{"department":"<anystring>","typeOfEmployee":"<anystring>","firstname":"<anystring>","lastname":"<anystring>","birthdate":"<anydate>","maxWorkHours":0,"username":"<anystring>","permissions":"<anystring>","lastUpdate":"<anydate>"}';
let employee1 = JSON.parse(employeeString);
console.log(employee1);

（操场上的代码）

如果您尝试JSON.parse在不是字符串的对象上使用它：

let e = {
    "department": "<anystring>",
    "typeOfEmployee": "<anystring>",
    "firstname": "<anystring>",
    "lastname": "<anystring>",
    "birthdate": "<anydate>",
    "maxWorkHours": 3,
    "username": "<anystring>",
    "permissions": "<anystring>",
    "lastUpdate": "<anydate>"
}
let employee2 = JSON.parse(e);

然后，您将收到错误消息，因为它不是字符串，而是对象，如果您已经以这种形式使用它，则无需使用JSON.parse。

但是，正如我所写的那样，如果您采用这种方式，那么您将没有类的实例，只有一个具有与类成员相同属性的对象。

如果您想要一个实例，那么：

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