我有一个复杂的对象,我想对其进行序列化和反序列化并获取相同类型的对象。
let workflow = new Workflow();
console.log(`workflow is instanceof Workflow: ${workflow instanceof Workflow}`);
console.log(workflow);
let json = JSON.stringify(workflow);
console.log(json);
let workflow2 = JSON.parse(json) as Workflow;
console.log(workflow2);
console.log(`workflow2 is instanceof Workflow: ${workflow2 instanceof Workflow}`);
let workflow3: Workflow = JSON.parse(json) as Workflow;
console.log(workflow3);
console.log(`workflow3 is instanceof Workflow: ${workflow3 instanceof Workflow}`);
控制台输出为:
是否有现成的解决方案,或者我需要手动重新实例化复杂对象并设置其所有属性?
最佳答案
您可以将对象的 fromJSON()
与 reviver
函数一起使用到 JSON.parse()
来实现您想要的效果。
例如:
type Serialized<T> = Pick<T, keyof T> & { _type: string };
class Workflow {
foo: number;
constructor(foo: number) {
this.foo = foo;
}
public toJSON(): Serialized<Workflow> {
return {
_type: this.constructor.name,
...this
};
}
public static fromJSON(source: Serialized<Workflow>): Workflow {
return new Workflow(source.foo);
}
}
function reviver(key: string, value: any): any {
if (typeof value === "object" && value && "_type" in value) {
switch (value._type) {
case "Workflow": return Workflow.fromJSON(value);
}
}
return value;
}
const w = new Workflow(42);
console.log(w instanceof Workflow);
const s = JSON.stringify(w);
console.log(s);
const w2 = JSON.parse(s, reviver) as Workflow;
console.log(w2.foo);
console.log(w2 instanceof Workflow);
打印:
true
{"_type":"Workflow","foo":42}
42
true
亲自尝试一下 playground !
关于javascript - 如何在 Typescript 中序列化/反序列化复杂对象,例如反序列化对象与序列化对象的类型相同,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57333334/