如何将 JSON 对象解析为 TypeScript 对象

推荐答案编译器允许您将 JSON.parse 返回的对象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 Employee 的实例，而是有一个具有相同属性的对象(如您在控制台中看到的那样).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.一个更简单的例子: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 errorslogA(a2);(操场上的代码)没有错误，因为a1满足A类型，因为它拥有它的所有属性，而且logA函数可以在没有运行时调用错误即使它收到的不是 A 的实例，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.当你的类是简单的数据对象并且没有方法时，这很有效，但是一旦你引入了方法，事情就会变得很糟糕: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(代码在操场上)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)如果您尝试在对象不是字符串时对它使用 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);然后你会得到错误，因为它不是一个字符串，它是一个对象，如果你已经有了这种形式的它，那么就没有必要使用 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，如果你采用这种方式，那么你将没有类的实例，只有一个与类成员具有相同属性的对象.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.如果你想要一个实例，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！ 上岸，阿里云！