Typescript 如何在省略一个参数的情况下从函数类型中提取参数类型

假设我有以下接口，它定义了一个函数:

export declare interface NavigationGuard {
    (to: RouteLocationNormalized, from: RouteLocationNormalized, next: NavigationGuardNext): NavigationGuardReturn | Promise<NavigationGuardReturn>;
}

我想以这种类型结束:

export SomeNavigationGuardType = (to: RouteLocationNormalized, from: RouteLocationNormalized): NavigationGuardReturn | Promise<NavigationGuardReturn>

利用最初的NavigationGuard型……

我try 使用实用程序类型Parameters，提取并使用Omit来删除下一个参数，但Parameters实用程序类型正在提取到元组.

我试着利用:

type NavigationGuardParams = Omit<Parameters<NavigationGuard>, 'next'>

export type UseRouterNivigationGuards = {
  beforeEach?: (arg: NavigationGuardParams) => ReturnType<NavigationGuard>
  beforeResolve?: (arg: NavigationGuardParams) => ReturnType<NavigationGuard>
  afterEach?: NavigationHookAfter
}

和

type NavigationGuardParams = Omit<Parameters<NavigationGuard>, 'next'>

export type UseRouterNivigationGuards = {
  beforeEach?: (...arg: NavigationGuardParams) => ReturnType<NavigationGuard>
  beforeResolve?: (...arg: NavigationGuardParams) => ReturnType<NavigationGuard>
  afterEach?: NavigationHookAfter
}

但这两个结果都会导致用法/参数错误，第一个结果如下:

"Expected 1 arguments, but got 2."

第二次try 给出:

"Argument of type '[RouteLocationNormalized, RouteLocationNormalized]' is not assignable to parameter of type 'NavigationGuardParams'."

因此，我请求帮助！有没有人有办法做到这一点？