How to use gorilla websocket and mux in the same application?

Question

func main() {
    router := mux.NewRouter().StrictSlash(true)
    router.HandleFunc("/api", home)
    fs := http.FileServer(http.Dir("../public"))
    http.Handle("/", fs)
    http.HandleFunc("/ws", handleConnections)
    go handleMessages()

    log.Println("http server started on :8000")
    err := http.ListenAndServe(":8000", nill)
    if err != nil {
        log.Fatal("ListenAndServe: ", err)
    }
}

With the above code, the /api route gives a 404. But if I change the err := http.ListenAndServe(":8000", nill) to err := http.ListenAndServe(":8000", router), the /api route works but the / route (which I serve the frontend) gives a 404.

How do I make them both work?

Edit: full code - https://codeshare.io/2Kpyb8

Answer 1

The second parameter type of the http.ListenAndServe function is http.Handler, if he is nil, http lib use http.DefaultServeMux http.Handler.

your /api route register to mux.NewRouter(), your / and /ws route register to http.DefaultServeMux,These are two different http.Handler objetc, You need to merge the routing requests registered by the two routers.

    router := mux.NewRouter().StrictSlash(true)
    router.HandleFunc("/api", home)
    // move ws up, prevent '/*' from covering '/ws' in not testing mux, httprouter has this bug.
    router.HandleFunc("/ws", handleConnections)
    // PathPrefix("/") match '/*' request
    router.PathPrefix("/").Handler(http.FileServer(http.Dir("../public")))
    go handleMessages()
    http.ListenAndServe(":8000", router)

gorilla/mux example not use http.HandleFunc function.