How to access params IF app is at the route outside of Outlet?

[danieljmann96]

Wondered if I could get some advice on how to set this app up.

So in my RootRoute I have the following setup.
I have a search page route that handles its own params.

<VariousProviders>
    <NavBar />
    <Outlet />
</VariousProviders>

The NavBar is as follows. When the search is submitted, the navigation to the search page works great with the params being set correctly. I want to autofill the search bar to get the search query from the params, but the page won't always be on "/search".

const { searchQuery } = useSearch({ from: searchRoute.id });
return (
    <VariousStuff />
    <SearchBar defaultValue={searchQuery} />
)

If I load the page on any other route, the error below appears

Invariant failed: useMatch("/search") is being called in a component that is meant to render the 'root' route. Did you mean to 'useMatch("/search", { strict: false })' or 'useRoute("/search")' instead?

Is there a best practice way of accessing params IF the route is "/search", without throwing an error?
If this isn't super clear happy to provide more details.

For reference the search route is as below

const searchRoute = new Route({
    getParentRoute: () => appRoot,
    path: '/search',
    component: lazyRouteComponent(() => import('../components/SearchPage'),
    validateSearch: (
        search: Partial<SearchUrl> & SearchSchemaInput
    ): SearchUrl | Record<string, never> => {
        const result = SearchPageUrlParamParser.safeParse(search);
        if (result.success) {
            return result.data
        }
        return defaultSearchValues;
    }
});

[danieljmann96]

I answered the question myself, but I will leave the answer below:

In my navigation bar, I now do the following:

const router = useRouter()
....
<SearchBar defaultValue={router.state.location.search.searchQuery ?? ''} />

The "router.state.location.search" will be a partial type of all of the different route's params combined; pretty nice 👍

[schiller-manuel]

Why not useSearch({ strict: false}) (hinted at by the error message)?

[danieljmann96]

Why not useSearch({ strict: false}) (hinted at by the error message)?

That gives a type error; it does still work when on the search route but also crashes my app when not on the search route.