从 Typescript 解析 JSON 恢复数据成员但不恢复类型：无法在结果上调用方法
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Parsing JSON from Typescript restores data members but not type: cannot call methods on result
提问by MarkusT
When I parse the JSON-stringified result of an object p1 back into another object p2, the second object gets the data associated with the first object, but I cannot call any nethods on it. Using http://www.typescriptlang.org/Playground/I tried the following:
当我将对象 p1 的 JSON 字符串化结果解析回另一个对象 p2 时，第二个对象获取与第一个对象关联的数据，但我无法在其上调用任何网络。使用http://www.typescriptlang.org/Playground/我尝试了以下操作：
class Person
{
constructor(public name: string, public age: number) {
}
Age() { return this.age; }
}
// Create a person
var p: Person = new Person("One", 1);
// Create a second person from the JSON representation
// of the first (NOTE: assert it is of type Person!)
var p2: Person = <Person>JSON.parse(JSON.stringify(p));
document.writeln("Start");
document.writeln(p.name); // OK: One
document.writeln(p.Age()); // OK: 1
document.writeln(p2.name); // OK: One
document.writeln(p2.age; // OK: 1
document.writeln(p2.Age()); // ERROR: no method Age() on Object
document.writeln("End");
How do I parse the JSON data and get a proper Person object?
如何解析 JSON 数据并获得正确的 Person 对象？
采纳答案by Nahuel Greco
try this:
试试这个：
// Create a person
var p: Person = new Person("One", 1);
// JSON roundtrip
var p_fromjson = JSON.parse(JSON.stringify(p))
// Hydrate it
var p2: Person = Object.create(Person.prototype);
Object.assign(p2, p_fromjson);
document.writeln(p2.Age()); // OK
回答by Fenton
JSON is a representation of the data only, not any behaviour.
JSON 只是数据的表示，而不是任何行为。
You could create a method on the object that accepts the JSON object and hydrates the data from it, but a JSON object cannot transfer the behaviour (methods etc) only the plain data.
您可以在接受 JSON 对象并从中提取数据的对象上创建一个方法，但 JSON 对象不能仅传输纯数据的行为（方法等）。
class Person
{
constructor(public name: string, public age: number) {
}
Age() { return this.age; }
static fromJson(json: string) {
var data = JSON.parse(json);
return new Person(data.name, data.age);
}
}
var p: Person = new Person("One", 53);
var jsonPerson = JSON.stringify(p);
var p2: Person = Person.fromJson(jsonPerson);
alert(p2.Age().toString());
