英文:
Routes returning 404 for mux gorilla
问题
在我的Go应用程序中，我正在使用gorilla/mux。
我希望实现以下功能：
http://host:3000/ 从子目录"frontend"静态地提供文件，并且
http://host:3000/api/ 及其子路径由指定的函数提供。
使用以下代码，两个调用都不起作用。
/index.html 是唯一一个起作用的（但它加载的资源不起作用）。我做错了什么？
package main
import (
"log"
"net/http"
"fmt"
"strconv"
"github.com/gorilla/mux"
)
func main() {
routineQuit := make(chan int)
router := mux.NewRouter().StrictSlash(true)
router.PathPrefix("/").Handler(http.FileServer(http.Dir("./frontend/")))
router.HandleFunc("/api", Index)
router.HandleFunc("/api/abc", AbcIndex)
router.HandleFunc("/api/abc/{id}", AbcShow)
http.Handle("/", router)
http.ListenAndServe(":" + strconv.Itoa(3000), router)
<- routineQuit
}
func Abc(w http.ResponseWriter, r *http.Request) {
fmt.Fprintln(w, "Index!")
}
func AbcIndex(w http.ResponseWriter, r *http.Request) {
fmt.Fprintln(w, "Todo Index!")
}
func AbcShow(w http.ResponseWriter, r *http.Request) {
vars := mux.Vars(r)
todoId := vars["todoId"]
fmt.Fprintln(w, "Todo show:", todoId)
}
英文:
In my Go application, I'm using gorilla/mux.
I would like to have
http://host:3000/ to be serving files statically from the subdirectory "frontend" and
http://host:3000/api/ and its subpaths being served by the specified functions.
With the following code, neither of the calls work.
/index.html is the only one that doesn (but not the resources being loaded by it). What am I doing wrong?
package main
import (
"log"
"net/http"
"fmt"
"strconv"
"github.com/gorilla/mux"
)
func main() {
routineQuit := make(chan int)
router := mux.NewRouter().StrictSlash(true)
router.PathPrefix("/").Handler(http.FileServer(http.Dir("./frontend/")))
router.HandleFunc("/api", Index)
router.HandleFunc("/api/abc", AbcIndex)
router.HandleFunc("/api/abc/{id}", AbcShow)
http.Handle("/", router)
http.ListenAndServe(":" + strconv.Itoa(3000), router)
<- routineQuit
}
func Abc(w http.ResponseWriter, r *http.Request) {
fmt.Fprintln(w, "Index!")
}
func AbcIndex(w http.ResponseWriter, r *http.Request) {
fmt.Fprintln(w, "Todo Index!")
}
func AbcShow(w http.ResponseWriter, r *http.Request) {
vars := mux.Vars(r)
todoId := vars["todoId"]
fmt.Fprintln(w, "Todo show:", todoId)
}
答案1
得分: 8
Gorilla的mux路由按照它们添加的顺序进行评估。因此，第一个匹配请求的路由将被使用。
在你的情况下，/处理程序将匹配每个传入的请求，然后在frontend/目录中查找文件，最后显示404错误。你只需要交换路由的顺序就可以运行它：
router := mux.NewRouter().StrictSlash(true)
router.HandleFunc("/api/abc/{id}", AbcShow)
router.HandleFunc("/api/abc", AbcIndex)
router.HandleFunc("/api", Abc)
router.PathPrefix("/").Handler(http.FileServer(http.Dir("./frontend/")))
http.Handle("/", router)
英文:
Gorilla's mux routes are evaluated in the order in which they are added. Therefore, the first route to match the request is used.
In your case, the / handler will match every incoming request, then look for the file in the frontend/ directory, then display a 404 error. You just need to swap your routes order to get it running:
router := mux.NewRouter().StrictSlash(true)
router.HandleFunc("/api/abc/{id}", AbcShow)
router.HandleFunc("/api/abc", AbcIndex)
router.HandleFunc("/api", Abc)
router.PathPrefix("/").Handler(http.FileServer(http.Dir("./frontend/")))
http.Handle("/", router)
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