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Auth Guarded Route doesn`t load lazy component
Nested Router Instances

Unable to upgrade from 1.17.4 to 1.17.5

Error when built
Flashing on refresh on a protected route
Route-based sidebar issues
/foo) that itself displays fullscreen. Upon clicking a link on the page, I attempt to navigate to /foo/$id/steps. This should open a side-panel (drawer) that overlaps the existing /foo route's content. Additionally, a link is available inside the side-panel to navigate from /foo/$id/steps –> /foo/$id/connect, which replaces a subset of the side-panel's contents with different contents.
Utilizing React Router V6, I was able to achieve this via roughly the following snippet
React Router V6...Redirecting with Pending Component
Wildcards
Minimal router tests type error
'__store.state' are incompatible between these types.
How can I adjust my minimal router to fit the type, or adjust the type to fit my minimal router?
...
Having a Slug instead of id in the route url
http://localhost:3000/topic/real-estate
http://localhost:3000/topic/real-estate
SSR Examples don't build?
frontend@0.0.0 build:client vite build src/entry-client.tsx --outDir dist/client...
Params typed incorrectly
routes/admin/$activityType/index.tsx, like so:
```
import { createFileRoute } from '@tanstack/react-router';
export const Route = createFileRoute('/admin/$activityType/')({...Implementing something similar to Processing Accumulated Route Context for just one app
Structuring navbar code, with an exception for the login page
__root.tsx file, since I thought "oh the navbar is essentially a layout for the entire app". However, from what I understood from the documentation, the root component is always rendered, so I can't prevent it from being rendered in a specific route such as the login.
Which leads to my question: should I keep my navbar code in the root component with a condition to only render when not in the login page? Or should I keep my root component mostly empty and create a new layout for the entirety of my app, with the exception of the login page?...How to return the last page without search params?
TypeError with Link component
`Route.fullPath` not accessible from lazy file route?
Back preserving previous searchParams
/users page that has pagination ie. /users?page=2. I then have a details page at /user/<id> eg /user/7. I want to mirror the functionality of the browser and preserve that search param from the previous page with a custom "back" button on-page.
I've read the docs but can't find a clear way to do this — preferably using <Link> if possible to keep it simple....Issue with Route.useParams() in a lazy file route
Link params type error
Object literal may only specify known properties, and 'id' does not exist in type '(current: {} | { category: string; } | {} | { id: string; } | { id: string; } | { category: string; id: string; }) => never'.ts(2353)
Object literal may only specify known properties, and 'id' does not exist in type '(current: {} | { category: string; } | {} | { id: string; } | { id: string; } | { category: string; id: string; }) => never'.ts(2353)