如何将JSON对象解析为TypeScript对象 [英] How to parse a JSON object to a TypeScript Object

问题描述

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.我在做什么错了?

I am currently trying to convert my received JSON Object into a TypeScript class with the same attributes and I cannot get it to work. What am I doing wrong?

员工类别

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);

推荐答案

之所以编译器允许您将从JSON.parse返回的对象强制转换为类，是因为

The reason that the compiler lets you cast the object returned from JSON.parse to a class is because typescript is based on structural subtyping.
You don't really have an instance of an Employee, you have an object (as you see in the console) which has the same properties.

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);

(没有错误，因为a1满足类型A，因为它具有所有属性，并且即使接收到的不是A的实例，也可以在没有运行时错误的情况下调用logA函数只要它具有相同的属性.

There's no error because a1 satisfies type A because it has all of its properties, and the logA function can be called with no runtime errors even if what it receives isn't an instance of A as long as it has the same properties.

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:

That works great when your classes are simple data objects and have no methods, but once you introduce methods then things tend to break:

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

(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);

(如果您尝试在不是字符串的对象上使用JSON.parse:

If you're trying to use JSON.parse on your object when it's not a string:

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.

Then you'll get the error because it's not a string, it's an object, and if you already have it in this form then there's no need to use JSON.parse.

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

But, as I wrote, if you're going with this way then you won't have an instance of the class, just an object that has the same properties as the class members.

如果您想要一个实例，那么:

If you want an instance then:

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>"
});

这篇关于如何将JSON对象解析为TypeScript对象的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！