TanStackT
TanStack•6mo ago•
3 replies
skinny-azure

How to infer the type of Route component with the context of it?

Let's say I have this component:

import { createFileRoute } from "@tanstack/react-router";

import { Stack, Title } from "@mantine/core";

import { ReturnReasonTable } from "./-components/table";

export const Route = createFileRoute("/(dashboard)/utilities/return-reason/")({
  loader: async ({ context: { queryClient, trpc } }) =>
    await queryClient.ensureQueryData(trpc.returnReason.getReturnReasons.queryOptions()),
  component: ReturnReasonPage,
  head: () => ({
    meta: [
      {
        title: "Return Reason",
      },
    ],
  }),
});

function ReturnReasonPage() {
  const { session } = Route.useRouteContext();

  return (
    <Stack>
      <Title order={4}>Return Reason</Title>

      <ReturnReasonTable session={session} />
    </Stack>
  );
}


I want to create a function that passes Route but I don't know how to infer the type of Route with the context in it (I have trpc, queryClient, session, and error).

I'm trying to create this component as a kind of pre process for ReturnReasonPage to check if the user has a permission.

import { redirect } from "@tanstack/react-router";

import { paths } from "#lib/paths.ts";

import { type Action, type Resource, checkPermission } from "@repo/shared/lib";

import type { Session } from "#lib/auth-client.ts";

export type WithAuthProps = {
  session: Session;
};

export function withAuth<T extends WithAuthProps>(
  Route: unknown,
  Component: React.ComponentType<T>,
  resource?: Resource,
  action?: Action
) {
  return async function AuthenticatedComponent(props: Omit<T, keyof WithAuthProps>) {
    const { session } = Route.useRouteContext();

    if (!session) {
      redirect({ to: paths.auth.signIn });
    }

    if (resource && action && !checkPermission({ resource, action, session })) {
      redirect({ to: "/401" });
    }

    return (
      <Component
        {...(props as T)}
        session={session}
      />
    );
  };
}
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