TypeScript:在构造函数中实现接口可能吗？

Question

我有以下界面:

interface SMJSPacket {
  header: {
    tag: string;
    method: string;
    type: string;
  };
  response?: {
    status: string;
    content: string;
  };
  event?: {
    key?: string;
    action?: string;
  };
  request?: {
    run?: string;
  };
}

然后我想将它作为一个类实现,并在构造函数中设置属性:

  class Request implements SMJSPacket {
    constructor(data: any, method: string) {
      this.header = {
        type: 'request',
        method: method || 'calld',
        tag: Request.getTag()
      }
      this.request = data;
    }
    static getTag(): string {
      return '_' + goog.now() + '_' + utils.getRandomBetween(1, 1000);
    }
  }

但是根据编译器Request没有实现接口.我不明白如何检查它,虽然它根据构造阶段的界面填充了一切,如果用JavaScript编写,这将工作正常,类型检查Closure工具中的相同的东西也完美.我的想法是,我想将接口实现为类,因此我可以在原型中使用实用程序方法,但仍然可以轻松地转换为JSON字符串.

有任何想法吗？

谢谢

Answer 1

语言服务将静态分析您的界面声明,并且因为您已经表示它需要您的header成员,它应该成为类声明的一部分:

class Request implements SMJSPacket {
    header: { tag: string; method: string; type: string; };

    constructor(data: any, method: string) {
        this.header = {
            type: "request",
            method: (method || "calld"),
            tag: Request.getTag()
        };
    }

    static getTag(): string {
        return "tag stuff";
    }
}

别担心,输出javascript更精简:

var Request = (function () {
    function Request(data, method) {
        this.header = {
            type: "request",
            method: (method || "calld"),
            tag: Request.getTag()
        };
    }
    Request.getTag = function getTag() {
        return "tag stuff";
    }
    return Request;
})();