Golang HTTP Mux更改处理程序功能

Question

I am fairly new to Go and have not been able to find any information on this, maybe it is just not possible at this time.

I am trying to delete or replace a mux route (using http.NewServeMux, or gorilla's mux.Router). My end goal is to be able to enable/disable a route or set of routes without having to restart the program.

I can probably accomplish this on a handler to handler basis and just return 404 if that feature is "disabled", but I would rather find a more general way to do this since I would like to implement it for every route in my application.

Or would I be better off just keeping track of disabled url patterns and using some middleware to prevent handler execution?

If someone can at least point me in the right direction, I will absolutely post code examples of a solution assuming there is one. Thanks!

Answer 1

There's no built in way, but it is easy enough to implement play.

type HasHandleFunc interface { //this is just so it would work for gorilla and http.ServerMux
    HandleFunc(pattern string, handler func(w http.ResponseWriter, req *http.Request))
}
type Handler struct {
    http.HandlerFunc
    Enabled bool
}
type Handlers map[string]*Handler

func (h Handlers) ServeHTTP(w http.ResponseWriter, r *http.Request) {
    path := r.URL.Path
    if handler, ok := h[path]; ok && handler.Enabled {
        handler.ServeHTTP(w, r)
    } else {
        http.Error(w, "Not Found", http.StatusNotFound)
    }
}

func (h Handlers) HandleFunc(mux HasHandleFunc, pattern string, handler http.HandlerFunc) {
    h[pattern] = &Handler{handler, true}
    mux.HandleFunc(pattern, h.ServeHTTP)
}

func main() {
    mux := http.NewServeMux()
    handlers := Handlers{}
    handlers.HandleFunc(mux, "/", func(w http.ResponseWriter, r *http.Request) {
        w.Write([]byte("this will show once"))
        handlers["/"].Enabled = false
    })
    http.Handle("/", mux)
    http.ListenAndServe(":9020", nil)
}