Golang logging http responses (in addition to requests)

go gorilla

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1922 Репутация автора

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 Golang 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?

Автор: Eric Broda Источник Размещён: 18.07.2016 06:42

Ответы (3)


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8732 Репутация автора

edit sorry, I didn't notice your mention of gorilla-mux, I have only tried this with gin, but if it uses middlewares this should still work.


the trick is, c.Next() in a middleware blocks until all subsequent middlewares return. Here's a logrus solution. Put this as your first middleware:

func Logrus(logger *logrus.Logger) gin.HandlerFunc {
    return func(c *gin.Context) {
        start := time.Now().UTC()
        path := c.Request.URL.Path
        c.Next()
        end := time.Now().UTC()
        latency := end.Sub(start)
        logger.WithFields(logrus.Fields{
            "status":     c.Writer.Status(),
            "method":     c.Request.Method,
            "path":       path,
            "ip":         c.ClientIP(),
            "duration":   latency,
            "user_agent": c.Request.UserAgent(),
        }).Info()
    }
}
GinEngine.Use(Logrus(logrus.StandardLogger()))
Автор: Plato Размещён: 18.07.2016 07:38

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1922 Репутация автора

Решение

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]
:
Автор: Eric Broda Размещён: 19.07.2016 09:51

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68 Репутация автора

The accepted answer by Eric Broda won't help much if you want to actually send your response to the client. I've made a modification to that code that will actually work:

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))        

        // this copies the recorded response to the response writer
        for k, v := range rec.HeaderMap {
            w.Header()[k] = v
        }
        w.WriteHeader(rec.Code)
        rec.Body.WriteTo(w)
    }
}
Автор: ouidevelop Размещён: 11.04.2018 01:47
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