golang资源被解释为样式表，但使用MIME类型text/plain传输

Question

我在使用golang开发网页时遇到问题。服务器文件(main.go):

package main
import (
"net/http"
"io/ioutil"
"strings"
"log"
)
type MyHandler struct {
}
func (this *MyHandler) ServeHTTP(w http.ResponseWriter, r *http.Request) {
path := r.URL.Path[1:]
log.Println(path)
data, err := ioutil.ReadFile(string(path))
if err == nil {
var contentType string
if strings.HasSuffix(path, ".css") {
contentType = "text/css"
} else if strings.HasSuffix(path, ".html") {
contentType = "text/html"
} else if strings.HasSuffix(path, ".js") {
contentType = "application/javascript"
} else if strings.HasSuffix(path, ".png") {
contentType = "image/png"
} else if strings.HasSuffix(path, ".svg") {
contentType = "image/svg+xml"
} else {
contentType = "text/plain"
}
w.Header().Add("Content Type", contentType)
w.Write(data)
} else {
w.WriteHeader(404)
w.Write([]byte("404 Mi amigo - " + http.StatusText(404)))
}
}
func main() {
http.Handle("/", new(MyHandler))
http.ListenAndServe(":8080", nil)
}

但是当我打字的时候http://localhost:8080/templates/home.html这是我看到的截图为什么我的页面没有正确加载？？我的css在哪里？？当我在main.go中处理内容类型时，为什么会出现错误"Resource explored as Stylesheet but transfer with MIME type text/plain:"？？

Answer 1

您的基本问题很简单：您需要Content-Type而不是Content Type。

但是，Go中有一种更好的方法可以将MIME类型与文件扩展名相匹配，特别是mime标准库包。我强烈建议你使用它。

Answer 2

如"Milo Christiansen"；mime在这种情况下会有所帮助。我还想指出，使用io.Copy而不是ioutil会更有效。ReadFile(如果流复制可以在内核级别发生，它会这样做)

package main
import (
"io"
"log"
"mime"
"net/http"
"os"
"path/filepath"
)
func main() {
http.HandleFunc("/", ServeFile)
log.Fatal(http.ListenAndServe(":8400", nil))
}
func ServeFile(resp http.ResponseWriter, req *http.Request) {
path := filepath.Join("./resource/", req.URL.Path)

file, err := os.Open(path)
if err != nil {
http.Error(resp, err.Error(), http.StatusNotFound)
return
}
if contentType := mime.TypeByExtension(filepath.Ext(path)); len(contentType) > 0 {
resp.Header().Add("Content-Type", contentType)
}
io.Copy(resp, file)
}