0

我一直在尝试用 go-chi 实现本教程，尤其是关于包装/将参数传递给包装器的部分。

我的目标是能够用带有自定义参数的中间件包装一些特定的路由，而不是让中间件对我的所有路由都是“全局的”，但我在做这件事时遇到了问题。

package main

import (
    "context"
    "io"
    "net/http"

    "github.com/go-chi/chi"
    "github.com/go-chi/chi/middleware"
)

func main() {

    r := chi.NewRouter()
    r.Use(middleware.Logger)

    r.Get("/user", MustParams(sayHello, "key", "auth"))

    http.ListenAndServe(":3000", r)
}


func sayHello(w http.ResponseWriter, r *http.Request) {
    w.Write([]byte("hi"))
}

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

        q := r.URL.Query()
        for _, param := range params {
            if len(q.Get(param)) == 0 {
                http.Error(w, "missing "+param, http.StatusBadRequest)
                return // exit early
            }
        }
        h.ServeHTTP(w, r) // all params present, proceed
    })
}

我越来越cannot use sayHello (type func(http.ResponseWriter, *http.Request)) as type http.Handler in argument to MustParams: func(http.ResponseWriter, *http.Request) does not implement http.Handler (missing ServeHTTP method)

如果我尝试输入断言它正在做r.Get("/user", MustParams(http.HandleFunc(sayHello), "key", "auth"))

我得到了错误cannot use MustParams(http.HandleFunc(sayHello), "key", "auth") (type http.Handler) as type http.HandlerFunc in argument to r.Get: need type assertion

我似乎无法找到使其工作的方法或能够用中间件包装单个路由。

4

1 回答 1

1

有一个函数 http.HandleFunc，然后有一个类型 http.HandlerFunc。您正在使用前者，但您需要后者。

r.Get("/user", MustParams(http.HandlerFunc(sayHello), "key", "auth"))
于 2021-01-12T04:20:30.757 回答