How to get the full URL from an Http4s Request-CodePudding

I'm trying to obtain the beginning of the URL that a client used to access a webserver in Http4s. I.e. given a Request[F] I want to obtain the string "http://localhost:8080" or "https://my-app.company.com" depending on where the server is deployed.

In a server implementation with Http4s, when receiving a Request via something like

HttpRoutes.of[IO] {
  case req =>
    println(req.uri)
    ???
}

I only ever seem to get a relative URI, e.g. / or /foo/bar - the scheme and authority fields on the Uri are both None.

It looks like I can pull the localhost:8080 part from the Host header on the request, but I have no solution for the scheme (http:// or https://).

How can I fill in the correct value for request.uri.scheme? And is there a more appropriate way to obtain the host?

FWIW I'm using http4s 0.22

CodePudding user response：

You can't get it while using Blaze as an HTTP server. There are workarounds. You can catch bound of a server:

By port:

[io-compute-4] INFO org.http4s.blaze.channel.nio1.NIO1SocketServerGroup - Service bound to address /[0:0:0:0:0:0:0:0]:8080

By schema:

[io-compute-4] INFO org.http4s.blaze.server.BlazeServerBuilder - http4s v1.0.0-M33 on blaze v1.0.0-M33 started at http://[::]:8080/

Or you can map 8080 to HTTP and 8081 to HTTPS and make pattern matching on it.

A bad workaround would be to add it as an HTTP header and read it via Headers It legacy practice but many legacy software companies are still using this approach.