React-Router-Dom 6 - How to dynamically render a component? - reactjs

Question

My old method:
<Route
key={i}
path={path}
render={(props) => {
if (!localStorage.getItem("token")) {
<Redirect
to={{ pathname: "/login", state: { from: props.location } }}
/>
}
return (
<AuthLayout>
<Component {...props} />
</AuthLayout>
);
}}
/>
Replacing render with the new element gives me:
Functions are not valid as a React child. This may happen if you return a Component instead of from render
Apparently the new API simply expects:
<Route
key={i}
path={path}
element={
<Component />
}
/>
What I'm really trying to accomplish is to dynamically render the component as such:
{authProtectedRoutes.map(({ path, Component }, i) => {
<Route
key={i}
path={path}
element={
// If no auth token, redirect to login
if (!token) {
<Navigate to="/login" />
} else {
<AuthLayout>
<Component />
</AuthLayout>
}
}
/>
})}
Not sure how to do this ...
EDIT:
My array of components is as such:
const authProtectedRoutes = [
{ path: "/dashboard", Component: Dashboard },
{ path: "/pages-starter", Component: StarterPage },
When I try to return Component in my loop I get:
React.jsx: type is invalid -- expected a string (for built-in
components) or a class/function (for composite components) but got:
undefined. You likely forgot to export your component from the file
it's defined in, or you might have mixed up default and named imports.

Answer 1

element={
// If no auth token, redirect to login
if (!token) {
<Navigate to="/login" />
} else {
<AuthLayout>
<Component />
</AuthLayout>
}
}
You can't do an if in the middle of jsx, but you can do a conditional operator:
element={!token ? (
<Navigate to="/login" />
) : (
<AuthLayout>
<Component />
</AuthLayout>
)}

Answer 2

The element prop expects a ReactNode (a.k.a. JSX) and not javascript (i.e. the if-statement).
Since it seems you render your authenticated routes in bulk a more optimal solution would be to wrap them all in a single AuthLayout component that checks the token. Instead of rendering the children prop it renders an Outlet for nested routes to be rendered into.
Example:
const AuthLayout = ({ token }) => {
// ... existing AuthLayout logic
return token
? (
<div /* awesome auth layout CSS style */>
...
<Outlet /> // <-- nested routes render here
</div>
)
: <Navigate to="/login" />;
};
Don't forget to return the Route from the map callback.
<Route element={<AuthLayout token={token} />}>
{authProtectedRoutes.map(({ path, Component }) => (
<Route key={path} path={path} element={<Component />} />
))}
</Route>

Answer 3

Nice routing-related question. First of all, I found useful code example from react-router-dom's github: https://github.com/remix-run/react-router/blob/2cd8266765925f8e4651d7caf42ebe60ec8e163a/examples/auth/src/App.tsx#L104
Here, instead of putting some logics inside "element" or "render" authors suggest to implement additional RequireAuth component and use it in routing setup like following:
<Route
path="/protected"
element={
<RequireAuth>
<SomePageComponent />
</RequireAuth>
}
....
This approach would allow to incapsulate auth-related checks inside this new RequireAuth component and as a result make your application routing setup more "lightweight"
As a "brief" example, I created following piece of code you could reference to:
function App() {
return (
<BrowserRouter>
<AppRoutes />
</BrowserRouter>
);
}
const RequireAuth = ({ children }) => {
const token = localStorage.getItem('token');
const currentUrl = useHref();
return !token ? (
<Navigate to={`/login?redirect=${currentUrl}`} />
) : (
children
);
};
const authProtectedRoutes = [
{ path: '/', component: PaymentsPage },
{ path: '/user', component: UserInfoPage },
];
const AppRoutes = () => (
<Routes>
{authProtectedRoutes.map((r) => (
<Route
key={r.path}
path={r.path}
element={
<RequireAuth>
<AuthLayout>
<r.component />
</AuthLayout>
</RequireAuth>
}
/>
))}
<Route path="/login" element={<LoginPage />} />
<Route path="*" element={<NotFoundPage />} />
</Routes>
);