http请求’a’进来,根据此请求中的POST有效负载启动 *** 作(在post或url中使用某种唯一标识符).由’a’启动的异步过程将使用原始唯一标识符(请求’b’)响应同一服务器,而请求’a’仍然打开.我想根据请求’b’响应将该响应传达回请求’a’.
解决方法 虽然可以通过频道完成此 *** 作,我更喜欢受互斥锁保护的哈希(map),
因为在这种情况下它更容易.
给你一个想法,让你去：
package mainimport ( "fmt" "net/http" "sync")type state struct { *sync.Mutex // inherits locking methods Vals map[string]string // map IDs to values}var State = &state{&sync.Mutex{},map[string]string{}}func get(rw http.ResponseWriter,req *http.Request) { State.Lock() defer State.Unlock() // ensure the lock is removed after leaving the the function ID := req.URL.query().Get("ID") // if you need other types,take a look at strconv package val := State.Vals[ID] delete(State.Vals,ID) rw.Write([]byte("got: " + val))}func post(rw http.ResponseWriter,req *http.Request) { State.Lock() defer State.Unlock() ID := req.FormValue("ID") State.Vals[ID] = req.FormValue("val") rw.Write([]byte("go to http://localhost:8080/?ID=42"))}var form = `<HTML> <body> <form action="/" method="POST"> ID: <input name="ID" value="42" /><br /> Val: <input name="val" /><br /> <input type="submit" value="submit"/> </form> </body></HTML>`func formHandler(rw http.ResponseWriter,req *http.Request) { rw.Write([]byte(form))}// for real routing take a look at gorilla/mux packagefunc handler(rw http.ResponseWriter,req *http.Request) { switch req.Method { case "POST": post(rw,req) case "GET": if req.URL.String() == "/form" { formHandler(rw,req) return } get(rw,req) }}func main() { fmt.Println("go to http://localhost:8080/form") // thats the default webserver of the net/http package,but you may // create custom servers as well err := http.ListenAndServe("localhost:8080",http.HandlerFunc(handler)) if err != nil { fmt.Println(err) }} 总结 以上是内存溢出为你收集整理的Golang：异步http服务器中的共享通信全部内容，希望文章能够帮你解决Golang：异步http服务器中的共享通信所遇到的程序开发问题。
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