Go http.FileServer gives unexpected 404 error-CodePudding

I'm trying to run two file servers, one of them serving index.html in the ui folder, and one other serving some other static files, like the code below:

package main

import (
    "log"
    "net/http"
)

func main() {
    srv := http.NewServeMux()

    // File server 1
    uiServer := http.FileServer(http.Dir("./ui"))
    srv.Handle("/", uiServer)

    // File server 2
    staticFilesServer := http.FileServer(http.Dir("./files"))
    srv.Handle("/files", staticFilesServer)

    if err := http.ListenAndServe(":8080", srv); err != nil {
        log.Fatal(err)
    }
}

Both fileServer objects are defined in the exact same way, and the first one (uiServer) works fine, but the second (staticFilesServer on localhost:8080/files), gives me 404.

I narrowed down the problem and removed the first one (working file server), just like the code below:

package main

import (
    "log"
    "net/http"
)

func main() {
    srv := http.NewServeMux()

    staticFilesServer := http.FileServer(http.Dir("./files"))
    srv.Handle("/files", staticFilesServer)

    if err := http.ListenAndServe(":8080", srv); err != nil {
        log.Fatal(err)
    }
}

But it still gives me 404 on the path localhost:8080/files

If I change handle path from /files to /, it works as expected, but that's not what i want, I just want to know is it even possible to serve on paths other than / and how can I achieve that.

Also, my folder structure:

|- main.go
|- ui
|--- index.html
|- files
|--- file1.txt
|--- file2.csv
|--- file3.img

CodePudding user response：

I realized that http.Dir() and http.ServeMux.Handle() would have a relation, they actually sum up their paths, as show below:

srv.Handle("/files/", http.FileServer(http.Dir("."))

Code above serves everything inside ./files folder, not . (as written in Dir(".")) And it fixed my problem.