我从远程睡觉服务器读取了一个json对象.此JSON对象具有TypeScript类的所有属性(根据设计).如何将接收到JSON对象转换为类型变量？
I don't want to populate a typescript var (ie have a constructor that takes this JSON object). It's large and copying everything across sub-object by sub-object & property by property would take a lot of time.
Update: You can however cast it to a typescript interface!
You can't simple cast a plain-old-JavaScript result from an Ajax request into a prototypical JavaScript/TypeScript class instance. There are a number of techniques for doing it, and generally involve copying data. Unless you create an instance of the class, it won't have any methods or properties. It will remain a simple JavaScript object.
如果只处理数据，则可以对接口执行强制转换(因为它纯粹是编译时 struct )，这需要使用一个TypeScript类，该类使用数据实例并对该数据执行操作.
复制数据的一些示例:
本质上，你只是:
var d = new MyRichObject();
d.copyInto(jsonResult);