I am using Go and the Gorilla web toolkit's mux and handler packages to build a complex application, part of which requires a http server. Gorilla's mux and handler packages work wonderfully and I am able to successfully get the http server up and running and it has been quite simple to log requests.
However, I am unable to determine how I may log responses. Ideally, I would like a mechanism, similar to Gorilla's LoggingHandler, that "wraps" the logging mechanism easily.
Is there a Go package that does easily wraps / logs responses? Is there a way to use Go or Gorilla's capabilities in this fashion that I have not considered?
Thanks for the great suggestions. I tried a few of the suggestions and landed on a rather simple solution that uses a minimalist wrapper. Here is the solution that worked for me (feel free to offer comments, or better yet, other solutions):
import (
"fmt"
"log"
"net/http"
"net/http/httptest"
"net/http/httputil"
"github.com/gorilla/mux"
)
:
func logHandler(fn http.HandlerFunc) http.HandlerFunc {
return func(w http.ResponseWriter, r *http.Request) {
x, err := httputil.DumpRequest(r, true)
if err != nil {
http.Error(w, fmt.Sprint(err), http.StatusInternalServerError)
return
}
log.Println(fmt.Sprintf("%q", x))
rec := httptest.NewRecorder()
fn(rec, r)
log.Println(fmt.Sprintf("%q", rec.Body))
}
}
func MessageHandler(w http.ResponseWriter, r *http.Request) {
fmt.Fprintln(w, "A message was received")
}
And the following code will use the aforementioned handler:
:
router := mux.NewRouter()
router.HandleFunc("/", logHandler(MessageHandler))
:
Output from the above code will be something along the lines of:
:
2016/07/20 14:44:29 "GET ... HTTP/1.1\r\nHost: localhost:8088\r\nAccept: */*\r\nUser-Agent: curl/7.43.0\r\n\r\n"
2016/07/20 14:44:29 ...[response body]
:
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