Unexpected 404 error with Go http.FileServer

Question content

I am trying to run two file servers, one of which is served in the ui folder index.html service, and another server provides some other static files, such as the following code:

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)
    }
}

The two fileserver objects are defined exactly the same way, the first one (uiserver) works fine, but the second one (staticfilesserver on localhost:8080/files) gives me a 404.

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

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 the handle path from /files to / it works as expected, but that's not what I want, I'm just wondering if it's possible in ## Serving on a path other than #/ and how I achieved this.

Also, my folder structure:

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

Correct answer

I realized that

http.dir() and http.servemux.handle() would There are relationships, they actually summarize their paths like this:

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

The above code provides all contents in the

./files folder instead of . (as written in dir(".")) It solved my problem.

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