How to extract the route into the method?

[Oodachi]

I want to place the route separately, I quickly checked the example, it will fail to compile.

pub fn app(db: DbConn) -> Route {
    route()
        .at("/hello/:name", get(hello))
        .with(RateLimitLayer::new(5, Duration::from_secs(30)).compat())
        .with(TimeoutLayer::new(Duration::from_secs(30)).compat())
        .with(AddData::new(db))
}

[sunli829]

Should return impl Endpoint, because with wraps the original Route type into another type.

pub fn app(db: DbConn) -> impl Endpoint {
    route()
        .at("/hello/:name", get(hello))
        .with(RateLimitLayer::new(5, Duration::from_secs(30)).compat())
        .with(TimeoutLayer::new(Duration::from_secs(30)).compat())
        .data(db) // similar to   .with(AddData::new(db))
}

[Oodachi]

Thanks!