Go: How to combine two (or more) http.ServeMux?

Question

Given that you have two instances of http.ServeMux, and you wish for them to be served at the same port number, like so:

    muxA, muxB http.ServeMu         


        

Answer 1

The SeverMux type is itself a http.Handler, therefore you can nest them easily. For example:

mux := NewServeMux()
mux.AddHandler("server1.com:8080/", muxA)
mux.AddHandler("server2.com:8080/", muxB)

I'm not quite sure what you mean exactly with "combining" them. If you want to try one handler first and then another one in case of 404, you might do it like this (untested):

mux := http.HandlerFunc(func(w http.ResponseWriter, r *http.Request) {
    rec := httptest.NewRecorder()
    rec.Body = &bytes.Buffer{}
    muxA.ServeHTTP(rec, r)
    if rec.Code == 404 {
        muxB.ServeHTTP(w, r)
        return
    }
    for key, val := range rec.HeaderMap {
        w.Header().Set(key, val)
    }
    w.WriteHeader(rec.Code)
    rec.Body.WriteTo(w)
})

This has obviously some disadvantages like storing the whole response in memory. Alternatively, if you don't mind calling your handler twice, you can also set rec.Body = nil, check just rec.Code and call muxA.ServeHTTP(w, r) again in case of success. But it's probably better to restructure your application so that the first approach (nested ServerMux) is sufficient.