绕过golang http处理程序

Question

Let's say I have code like this

handler := middleware1.New(
    middleware2.New(
        middleware3.New(
            middleware4.New(
                NewHandler()
            ),
        ),
    ),
)
http.ListenAndServe(":8080", handler)

where handler has tons of middleware.

Now I want to create custom endpoint, which will skip all the middleware, so nothing what's inside serveHTTP() functions is executed:

http.HandleFunc("/testing", func(
    w http.ResponseWriter,
    r *http.Request,
) {
    fmt.Fprintf(w, "it works!")
    return
})
http.ListenAndServe(":8080", handler)

But this doesn't work and /testing is never reached. Ideally, I don't want to modify handler at all, is that possible?

Answer 1

You can use an http.ServeMux to route requests to the correct handler:

m := http.NewServeMux()
m.HandleFunc("/testing", func(w http.ResponseWriter, r *http.Request) {
    fmt.Fprintf(w, "it works!")
    return
})
m.Handle("/", handler)

http.ListenAndServe(":8080", m)

Using the http.HandleFunc and http.Handle functions will accomplish the same result using the http.DefaultServerMux , in which case you would leave the handler argument to ListenAndServe as nil .

Answer 2

try this, ListenAndServe handler is usually nil.

http.Handle("/", handler)

http.HandleFunc("/testing", func(
    w http.ResponseWriter,
    r *http.Request,
) {
    fmt.Fprintf(w, "it works!")
    return
})
http.ListenAndServe(":8080", nil)