英文:
How do I get the URL before routing?
问题
我想在路由之前获取路由的URL或名称，例如：
router.HandleFunc("/person", person.HandlePerson)
router.HandleFunc("/animal", animal.HandleAnimal)
我想在路由之前像这样知道路由的名称：
nameOfRoute := //我想让这个变量存储路由的名称（person或animal）
fmt.Println(nameOfRoute) //这行代码将打印路由的名称（"/animal"或"/person"）
router.HandleFunc("/person", person.HandlePerson)
router.HandleFunc("/animal", animal.HandleAnimal)
在Golang中，你可以通过使用http.HandlerFunc类型的函数来实现这一点。你可以创建一个自定义的HandlerFunc函数，它会在路由之前打印路由的名称，并将其存储在变量中。以下是一个示例代码：
package main
import (
"fmt"
"net/http"
)
func main() {
router := http.NewServeMux()
router.HandleFunc("/person", logAndHandle(person.HandlePerson, "/person"))
router.HandleFunc("/animal", logAndHandle(animal.HandleAnimal, "/animal"))
// 启动服务器
http.ListenAndServe(":8080", router)
}
func logAndHandle(handler http.HandlerFunc, routeName string) http.HandlerFunc {
return func(w http.ResponseWriter, r *http.Request) {
fmt.Println(routeName)
handler(w, r)
}
}
func HandlePerson(w http.ResponseWriter, r *http.Request) {
// 处理/person路由的逻辑
}
func HandleAnimal(w http.ResponseWriter, r *http.Request) {
// 处理/animal路由的逻辑
}
在上面的示例中，我们创建了一个名为logAndHandle的辅助函数，它接受一个HandlerFunc和路由名称作为参数。该函数会在路由之前打印路由的名称，并调用原始的HandlerFunc来处理请求。你可以根据需要修改logAndHandle函数的实现来满足你的需求。
英文:
I want to get the url or the neme of the route before routing it, ex:
router.HandleFunc("/person", person.HandlePerson)
router.HandleFunc("/animal", animal.HandleAnimal)
I want to know the route before routing like this:
nameOfRoute:= //I want this variable to store the name of the route (person or animal) before routing
fmt.Println(nameOfRoute) //This line will print the name of the route ("/animal", "/person")
router.HandleFunc("/person", person.HandlePerson)
router.HandleFunc("/animal", animal.HandleAnimal)
How can I do this in Golang
答案1
得分: 1
你不能以你展示的方式进行路由。那是因为这些语句
router.HandleFunc("/person", person.HandlePerson)
不是实际进行路由的操作，而是指示路由器如何进行路由的指令。
路由本身是通过路由器的一个“入口”函数来处理的。
然而，你可以使用多个路由器进行“级联”：
- 创建一个处理
/的路由器，然后创建另一个处理所有个别路由的路由器 - 就像目前所做的那样。 - 让第一个 - 根 - 路由器的唯一处理函数打印URL，然后仅调用内部路由器的入口函数。
我不知道你使用的是哪个路由器包（有很多，包括默认的），所以我不能告诉你那个函数是什么；在默认的路由器中，它是ServeHTTP。
请考虑将此作为作业来解决
使用默认的net/http的ServeMux，你甚至不需要一个根路由器，因为一个函数就足够了 - 就像这样：
mux := http.NewServeMux()
mux.Handle("/person", whatever)
...
server := http.Server{
Handler: func(w http.ResponseWriter, r *http.Request) {
fmt.Println(r.URL)
mux.ServeHTTP(w, r)
},
}
server.ListenAndServe()
英文:
You cannot — in the sense that you showed. That's because the statements
router.HandleFunc("/person", person.HandlePerson)
are not actions which carry out actual routing but rather directives to the router about how it should do routing.
The routing itself is handler via a single "entry" function of your router.
What you can do, however, is to employ "cascading" of several routers:
- Create a router which handles
/, and then create another router which handles all your indiviual routes — as it's currently done. - Make the sole handler function of the first — root — router print the URL and then merely call the inner's router entry function.
I have no idea which router package are you using (there are gazillions of them, including the stock one) so I cannot tell you what that function is; in a stock router it'sServeHTTP.
Please consider figuring this out as a homework
With a stock net/http's ServeMux, you do not even need a root router because a single function would do — something like this:
mux := http.NewServeMux()
mux.Handle("/person", whatever)
...
server := http.Server{
Handler: func(w http.ResponseWriter, r *http.Request) {
fmt.Println(r.URL)
mux.ServeHTTP(w, r)
},
}
server.ListenAndServe()
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