javascript - 将类的静态方法转换为变量 typescript

Question

大家好，目前我正在为我正在创建的应用程序/库提供服务，到目前为止，这是我可以弄清楚如何创建服务的最佳方式。理想情况下，我想将此方法更改为变量以便于阅读

import { NavigationRoute, NavigationParams } from 'react-navigation';
import { NavigationStackProp } from 'react-navigation-stack';

type Navigation =
  | NavigationStackProp<NavigationRoute<NavigationParams>, NavigationParams>
  | undefined;

export default class NavigationService {
  private static _navigator: Navigation;

  public static setNavigator(navigatorRef: Navigation) {
    this._navigator = navigatorRef;
  }

  public static navigate = (): Navigation => {
    // TODO: look into how to make this a variable
    return NavigationService._navigator;
  };
}

目前我有这个

      IconPress: () => NavigationService.navigate()?.openDrawer()

但希望它读起来像这样

      IconPress: () => NavigationService.navigate?.openDrawer()

Answer 1

这称为 getter :

export default class NavigationService {

    public static get navigate(): Navigation {
        return NavigationService._navigator;
    } 
}