我使用 Gorilla Mux 作為我的網絡服務器。我定義了一堆路由，但如果沒有匹配的路由，我希望它為我的 index.html 文件提供服務。func (mgr *ApiMgr) InstantiateRestRtr() *mux.Router { mgr.pRestRtr = mux.NewRouter().StrictSlash(true) mgr.pRestRtr.PathPrefix("/api/").Handler(http.StripPrefix("/api/", http.FileServer(http.Dir(mgr.fullPath+"/docsui")))) for _, route := range mgr.restRoutes { var handler http.Handler handler = Logger(route.HandlerFunc, route.Name) mgr.pRestRtr.Methods(route.Method) .Path(route.Pattern) .Name(route.Name) .Handler(handler) } // Here I want to catch any unmatched routes, and serve my '/site/index.html` file. // How can I do this? return mgr.pRestRtr}
1 回答
慕村225694
TA貢獻1880條經驗 獲得超4個贊
func NewRouter() *mux.Router {
router := mux.NewRouter().StrictSlash(true)
routes := client.GetRoutes()
for _, route := range routes {
var handler http.Handler
handler = route.HandlerFunc
handler = context.ClearHandler(handler)
router.
Methods(route.Method).
Path(route.Pattern).
Name(route.Name).
Handler(handler)
}
r := OGRouter{router: router}
router = r.router
router.NotFoundHandler = http.HandlerFunc(r.NotFoundHandler)
return router
}
/* Handled mux router drawback of giving 404 in every case i.e if either URI or METHOD
* is not matched mux router will give 404. So, this case is handled her, if Method will
* not match we will through 405
*/
func (router *OGRouter) NotFoundHandler(rw http.ResponseWriter, req *http.Request) {
r := *router.router
route := r.Get(req.RequestURI)
if route != nil {
if req.RequestURI == route.GetName() {
routeMatch := mux.RouteMatch{Route: route}
if !route.Match(req, &routeMatch) {
rw.WriteHeader(http.StatusMethodNotAllowed)
}
} else {
rw.WriteHeader(http.StatusNotFound)
}
} else {
rw.WriteHeader(http.StatusNotFound)
}
}
type OGRouter struct {
router *mux.Router
}
在您獲得 404 的地方，您可以提供一個新頁面
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