http - 将 http.NewServeMux 放入 http.NewServeMux

Question

这是我的代码。我想将 mux1 作为子路由器放入 mux.Handle 中。接下来，我运行代码。我可以访问路径/index 但我无法访问路径/index/sub1。不知道为什么我可以访问/index但不能访问/index/sub1？

package main

import (
    "io"
    "net/http"
)

func main() {
    mux1 := http.NewServeMux()
    mux1.HandleFunc("/", func(w http.ResponseWriter, r *http.Request) {
        io.WriteString(w, "sub index")
    })
    mux1.HandleFunc("/sub1", func(w http.ResponseWriter, r *http.Request) {
        io.WriteString(w, "sub 1")
    })

    mux := http.NewServeMux()
    mux.Handle("/index", mux1)

    http.ListenAndServe(":8000", mux)
}

Answer 1

您的示例不起作用，因为您使用了 /index 路径来注册“外部”处理程序，这是一个单一的路径(因为它没有以斜线结尾 /) 而不是根子树(即所有以 /index/ 开头的路径)。这记录在 http.ServeMux :

Patterns name fixed, rooted paths, like "/favicon.ico", or rooted subtrees, like "/images/" (note the trailing slash). Longer patterns take precedence over shorter ones, so that if there are handlers registered for both "/images/" and "/images/thumbnails/", the latter handler will be called for paths beginning "/images/thumbnails/" and the former will receive requests for any other paths in the "/images/" subtree.

而且因为“子路由器”不会看到您用来注册处理程序的路径，例如 "/" 或 "/sub1"，但是 完整路径，即:"/index/" 和"/index/sub1"。

所以主路由器应该做的是“剥离”“/index” 前缀，这是它注册到的路径(没有尾部斜线)。幸运的是，标准库有一个现成的解决方案:http.StripPrefix() .

所以工作解决方案:

mux1 := http.NewServeMux()
mux1.HandleFunc("/", func(w http.ResponseWriter, r *http.Request) {
    io.WriteString(w, "sub index")
})
mux1.HandleFunc("/sub1", func(w http.ResponseWriter, r *http.Request) {
    io.WriteString(w, "sub 1")
})

mux := http.NewServeMux()
mux.Handle("/index/", http.StripPrefix("/index", mux1))

http.ListenAndServe(":8000", mux)

测试它:

• 网址:http://localhost:8000/index/
  响应:子索引

• 网址:http://localhost:8000/index/sub1
  响应:sub 1