How to extract parameter types from a function type while omitting one parameter?

Let’s say I have the following interface which defines a function:

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

I’d like to end up with this type:

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

Utilising the original NavigationGuard type …

I have tried utilising the utility type Parameters, extracting and using Omit to remove the next param, but the Parameters utility type is extracting to a tuple.

I’ve tried utilising:

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

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

and

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

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

But both results in a usage / param error, the first giving this:

"Expected 1 arguments, but got 2."

The second attempt giving:

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

I’m therefore asking for help! Would anyone have a trick to be able to achieve this?

>Solution :

type SomeNavigationGuardType = Parameters<NavigationGuard> extends [...infer T, next: any] ? T : never