go - 扩展 HTTP 处理程序

Question

我的 Go 网络应用程序中有一个相当快速和肮脏的错误处理程序，它会引发 HTTP 错误，记录响应的重要部分并提供错误模板。我想删除我在处理程序中多次编写类似内容的重复:

err := doSomething()
if err != nil {
    serverError(w, r, err, code)
}

我很好地阅读了 Error Handling and Go文章涵盖了定义返回错误类型/结构的自定义 HTTP 处理程序类型(或者甚至返回 int, err 而不是):

type appHandler func(http.ResponseWriter, *http.Request) *appError

type appError struct {
        code int
        Err error
}

// Ensures appHandler satisfies the http.Handler interface
func (fn appHandler) ServeHTTP(w http.ResponseWriter, r *http.Request) {
    if err := fn(w, r); err != nil {
        switch err.Code {

        case http.StatusNotFound:
            http.NotFound(w, r)

        case http.StatusInternalServerError:
            http.Error(w, "message", http.StatusInternalServerError)

        default:
            http.Error(w, "message", err.Code)

        }
    }
}

但我不确定如何保留允许我像这样链接中间件的现有中间件功能/包装器:r.HandleFunc("/route", use(myHandler, middleware1, middleware2)) 其中 use 和我的中间件如下所示:

func use(h http.HandlerFunc, middleware ...func(http.HandlerFunc) http.HandlerFunc) http.HandlerFunc {
    for _, m := range middleware {
        h = m(h)
    }

    return h
}

 func AntiCSRF(h http.HandlerFunc) http.HandlerFunc {
        return http.HandlerFunc(func(w http.ResponseWriter, r *http.Request) {
                // do something
                // h.ServeHTTP(w,r)
            }
     }

据我所知，它类似于下面的内容(这是行不通的)。我收到一条错误消息，提示 cannot use m(h) (type http.Handler) as type appHandler in assignment: need type assertion。我该如何解决这个问题，同时仍然保持中间件本身“原样”？

您可以在这里找到一个(简化的) Playground 示例:http://play.golang.org/p/Cmmo-wK2Af

r.Handle("/route", use(myHandler, middleware.NoCache)) // Contrived example!

func use(h myHandlerType?, middleware ...func(http.Handler) http.Handler) http.Handler {
    for _, m := range middleware {
        h = m(h)
    }

    return h
}

func myHandler(w http.ResponseWriter, r *http.Request) *appError {

    // Extremely contrived example
    name := "Matt"
    _, err := fmt.Fprintf(w, "Hi %s", name)
    if err != nil {
        return &appError{500, err}
    }

    return nil
}

func contrivedMiddleware(h http.Handler) http.Handler {
    return http.HandlerFunc(func(w http.ResponseWriter, r *http.Request) {

        w.Header().Set("Cache-Control", "max-age=0, private, must-revalidate")
        w.Header().Set("X-Accel-Expires", "0")

        h.ServeHTTP(w, r)
    })
}

我错过了什么，是否有更好的方法来做到这一点？

Answer 1

感谢 #go-nuts 上“cronos”的帮助，我设法解决了这个问题。

该解决方案允许我使用自定义处理程序类型、链接中间件并避免重复包装处理程序(即 appHandler(myHandler))、中间件...:

type appHandler func(http.ResponseWriter, *http.Request) *appError

type appError struct {
    Code  int
    Error error
}

func (fn appHandler) ServeHTTP(w http.ResponseWriter, r *http.Request) {
    if e := fn(w, r); e != nil {

        switch e.Code {

        case http.StatusNotFound:
            notFound(w, r)
        case http.StatusInternalServerError:
            serverError(w, r, e.Error, e.Code)
        default:
            serverError(w, r, e.Error, e.Code)
        }
    }
}

func use(h appHandler, middleware ...func(http.Handler) http.Handler) http.Handler {
    var res http.Handler = h
    for _, m := range middleware {
        res = m(res)
    }

    return res
}

func someMiddleware(h http.Handler) http.Handler {
    return http.HandlerFunc(func(w http.ResponseWriter, r *http.Request) {

        w.Header().Set("Cache-Control", "max-age=0, private, must-revalidate")
        w.Header().Set("X-Accel-Expires", "0")
        h.ServeHTTP(w, r)
    })
}

func myHandler(w http.ResponseWriter, r *http.Request) *appError {

    err := doSomething()
    if err != nil {
        return &appError{500, err}
    }

    // render your template, etc.
    return nil
}

路由看起来像这样:r.Handle("/route", use(myHandler, someMiddleware))

显然，您可以修改 appHandler 以返回您喜欢的任何内容，向 appError 添加额外的字段等等。如果您想将中间件应用于所有路由，您的中间件也可以包装您的路由器 - 即 http.Handle("/", someMiddleware(r))