Question

我正在学习Go，并且正在编写一个简单的Web服务器，它使用通道来限制并发请求的数量。服务器在控制台上打印日志条目，显示它接收请求并处理它们，但客户端浏览器不显示任何输出。我尝试添加响应编写器的同花顺，但没有帮助。

作为一个菜鸟，我错过了什么？感谢您的任何提示/指示。

以下是代码：

package main

import (
    "fmt"
    "html"
    "net/http"
    "time"
)

// define a type to be used with our request channel
type clientRequest struct {
    r *http.Request
    w http.ResponseWriter
}

const (
    MaxRequests int = 10
)

// the request channel, to limit the number of simultaneous requests being processed
var reqChannel chan *clientRequest

func init() {
    reqChannel = make(chan *clientRequest, MaxRequests)
}

func main() {
    // create the server's handler
    var ServeMux = http.NewServeMux()
    ServeMux.HandleFunc("/", serveHandler)

    // start pool of request handlers, all reading from the same channel
    for i := 0; i < MaxRequests; i++ {
        go processRequest(i)
    }

    // create the server object
    s := &http.Server{
        Addr:           ":8080",
        Handler:        ServeMux,         // handler to invoke, http.DefaultServeMux if nil
        ReadTimeout:    10 * time.Second, // maximum duration before timing out read of the request
        WriteTimeout:   10 * time.Second, // maximum duration before timing out write of the response
        MaxHeaderBytes: 1 << 20,          // maximum size of request headers, 1048576 bytes
    }

    // start the server
    err := s.ListenAndServe()
    if err != nil {
        fmt.Println("Server failed to start: ", err)
    }
}

func serveHandler(w http.ResponseWriter, r *http.Request) {
    var newRequest = new(clientRequest)
    newRequest.r = r
    newRequest.w = w

    reqChannel <- newRequest // send the new request to the request channel
    fmt.Printf("Sent request to reqChannel for URL: %q\n", html.EscapeString(r.URL.Path))
}

func processRequest(instanceNbr int) {
    fmt.Printf("processRequest started for instance #%d\n", instanceNbr)
    for theRequest := range reqChannel { // receive requests from the channel until it is closed
        fmt.Printf("Got request from reqChannel for URL: %q\n", html.EscapeString(theRequest.r.URL.Path))

        // xxx this isn't working:
        fmt.Fprintf(theRequest.w, "processRequest instance #%d: URL is %q", instanceNbr, html.EscapeString(theRequest.r.URL.Path))
        if f, ok := theRequest.w.(http.Flusher); ok {
            f.Flush()
        }
    }
}

Answer 1

服务器在serveHandler返回时关闭响应。

一个解决方法是在处理请求之前阻止serveHandler。在以下代码中，工作程序关闭done以表示请求已完成。处理程序等待done关闭。

type clientRequest struct {
    r *http.Request
    w http.ResponseWriter
    done chan struct{}  // <-- add this line
}

func serveHandler(w http.ResponseWriter, r *http.Request) {
   var newRequest = new(clientRequest)
   newRequest.r = r
   newRequest.w = w
   newRequest.done = make(chan struct{})

   reqChannel <- newRequest // send the new request to the request channel
   fmt.Printf("Sent request to reqChannel for URL: %q\n", html.EscapeString(r.URL.Path))
   <-newRequest.done  // wait for worker goroutine to complete
}

func processRequest(instanceNbr int) {
   fmt.Printf("processRequest started for instance #%d\n", instanceNbr)
   for theRequest := range reqChannel { // receive requests from the channel until it is closed
       fmt.Printf("Got request from reqChannel for URL: %q\n", html.EscapeString(theRequest.r.URL.Path))

       fmt.Fprintf(theRequest.w, "processRequest instance #%d: URL is %q", instanceNbr, html.EscapeString(theRequest.r.URL.Path))
       if f, ok := theRequest.w.(http.Flusher); ok {
           f.Flush()
       }
       close(theRequest.done)  // signal handler that request is complete
   }
}

如果目标是限制活动处理程序的数量，那么您可以使用通道作为计数信号量来限制活动处理程序goroutine的数量：

var reqChannel = make(chan struct{}, MaxRequests)

func serveHandler(w http.ResponseWriter, r *http.Request) {
    reqChannel <- struct{}{} 
    // handle the request
    <-reqChannel
}

请注意，服务器在每个连接goroutine中运行处理程序。

更简单的是只写一个处理程序。大多数服务器不需要限制请求处理程序并发。

Answer 2

您的回答是part of the net/http code：

    // HTTP cannot have multiple simultaneous active requests.[*]
    // Until the server replies to this request, it can't read another,
    // so we might as well run the handler in this goroutine.
    // [*] Not strictly true: HTTP pipelining.  We could let them all process
    // in parallel even if their responses need to be serialized.
    serverHandler{c.server}.ServeHTTP(w, w.req)
    if c.hijacked() {
        return
    }
    w.finishRequest()

ServeHTTP返回后，请求完成。

所以你有一些解决方案：

放弃您的工作模式并在serveHandler
在完成serveHandler之前等待请求完全处理，具体如下：

（在我当地测试）

type clientRequest struct {
    r *http.Request
    w http.ResponseWriter
    done chan struct{}
}

func serveHandler(w http.ResponseWriter, r *http.Request) {
    var newRequest = new(clientRequest)
    newRequest.r = r
    newRequest.w = w
    newRequest.done = make(chan struct{})

    reqChannel <- newRequest // send the new request to the request channel
    fmt.Printf("Sent request to reqChannel for URL: %q\n", html.EscapeString(r.URL.Path))
    <-newRequest.done // wait for the worker to finish
}

func processRequest(instanceNbr int) {
    fmt.Printf("processRequest started for instance #%d\n", instanceNbr)
    for theRequest := range reqChannel { // receive requests from the channel until it is closed
        fmt.Printf("Got request from reqChannel for URL: %q\n", html.EscapeString(theRequest.r.URL.Path))

        // xxx this isn't working:
        fmt.Fprintf(theRequest.w, "processRequest instance #%d: URL is %q", instanceNbr, html.EscapeString(theRequest.r.URL.Path))
        if f, ok := theRequest.w.(http.Flusher); ok {
            f.Flush()
        }
        theRequest.done <- struct{}{}
    }
}

简单的Go Web服务器，没有在客户端看到响应

2 个答案: