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Styling of Links for the Currently Active Route

The same strategy you use to conditionally apply a CSS class to an element dependent on state will also be suitable for changing the styling of a link if it matches the current route. Type Route provides no specific API for this. Typically links in the navigation header of your application will have certain styling applied to them when they match the current route. For example:

type Props = {
  route: Route<typeof routes>
}

function Navigation({ route }: Props) {
  return <nav>
    <a {...routes.home().link} className={route.name === "home" ? "active" : undefined}>Home</a>
    <a {...routes.other().link} className={route.name === "other" ? "active" : undefined}>Other</a>
  </nav>
}

A light abstraction on top of this may look something like the following:

type LinkProps = {
  to: Route<typeof routes>;
  children?: React.ReactNode;
}

function Link({ to, children }: Props) {
  const route = useRoute();

  return <a
    {...to.link}
    className={to.name === route.name ? "active" : undefined}
  >
    {children}
  </a>;
}

function Navigation() {
  return <nav>
    <Link to={routes.home()}>Home</Link>
    <Link to={routes.other()}>Other</Link>
  </nav>
}

Related pages:

• Rendering Links
• Custom Link Behavior
• preventDefaultLinkClickBehavior