Question

给出以下代码：

interface TaskInterface {
  execute(): Promise<any>;
}

class Task implements TaskInterface {
  private config: any = {};
  public constructor(config: any) {
    this.config = Object.assign({}, config);
  }
  public execute(): Promise<any> {
    return Promise.resolve(`Hi, Mom ${this.config.info}`);
  }
}

class Pop extends Task {
  public constructor(config: any) {
    super(config);
  }
}
class Fetch extends Task {}

const taskData = { type: "pop" };
let TaskClass: Task;
if (taskData.type === "pop") TaskClass = Pop;
if (taskData.type === "fetch") TaskClass = Fetch;
const configuration = { settings: { power: "full" } };
if (TaskClass) {
  const task = new TaskClass(configuration.settings);
  task.execute();
}

我从TypeScript中收到以下错误：

src/test.ts:24:42 - error TS2739: Type 'typeof Pop' is missing the following properties from type 'Task': config, execute
24 if (taskData.type === "pop") TaskClass = Pop;
    Did you mean to use 'new' with this expression?

src/test.ts:25:44 - error TS2739: Type 'typeof Fetch' is missing the following properties from type 'Task': config, execute
25 if (taskData.type === "fetch") TaskClass = Fetch;
  Did you mean to use 'new' with this expression?

src/test.ts:28:16 - error TS2351: Cannot use 'new' with an expression whose type lacks a call or construct signature.
28   const task = new TaskClass(configuration.settings);
                  ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

我想知道什么是能够实现此目标的正确方法？这是an online working example：

Answer 1

类名是实例类型，要获取类本身的类型，您需要typeof className：

const taskData = { type: "pop" };
let TaskClass: typeof Task;
if (taskData.type === "pop") TaskClass = Pop;
if (taskData.type === "fetch") TaskClass = Fetch;
const configuration = { settings: { power: "full" } };
if (TaskClass) {
    const task = new TaskClass(configuration.settings);
    task.execute();
}

尝试使用子类时，获取“ Type'typeof <Class>'类型缺少类型的以下属性”错误

1 个答案: