Question

该程序同时产生了许多goroutine（getStock），我相信这会导致远程服务器立即断开连接。我不是想创建一个DOS，但我仍然希望积极地获取数据而不会出现“连接重置”错误。

哪些策略最多只能有N个（例如20个）同时连接？ golang Http客户端中是否有GET请求的内置队列？我还在学习，如果这类代码有更好的设计模式，那就太棒了。

输出

$ go run s1w.go 
sl(size): 1280
body: "AAPL",17.92
body: "GOOG",32.13
body: "FB",42.02
body: "AMZN",195.83
body: "GOOG",32.13
body: "AMZN",195.83
body: "GOOG",32.13
body: "FB",42.02
body: "AAPL",17.92
2017/07/26 00:01:23 NFLX: Get http://goanuj.freeshell.org/go/NFLX.txt: read tcp 192.168.86.28:56674->205.166.94.30:80: read: connection reset by peer
2017/07/26 00:01:23 AAPL: Get http://goanuj.freeshell.org/go/AAPL.txt: read tcp 192.168.86.28:56574->205.166.94.30:80: read: connection reset by peer
2017/07/26 00:01:23 NFLX: Get http://goanuj.freeshell.org/go/NFLX.txt: read tcp 192.168.86.28:56760->205.166.94.30:80: read: connection reset by peer
2017/07/26 00:01:23 FB: Get http://goanuj.freeshell.org/go/FB.txt: read tcp 192.168.86.28:56688->205.166.94.30:80: read: connection reset by peer
2017/07/26 00:01:23 AMZN: Get http://goanuj.freeshell.org/go/AMZN.txt: read tcp 192.168.86.28:56689->205.166.94.30:80: read: connection reset by peer
2017/07/26 00:01:23 AAPL: Get http://goanuj.freeshell.org/go/AAPL.txt: read tcp 192.168.86.28:56702->205.166.94.30:80: read: connection reset by peer

s1.go

package main
import (
        "fmt"
        "io/ioutil"
        "log"
        "net/http"
        "time"
)

// https://www.youtube.com/watch?v=f6kdp27TYZs (15m)
// Generator: function that returns a channel
func getStocks(sl []string) <-chan string {
        c := make(chan string)
        for _, s := range sl {
                go getStock(s, c)
        }
        return c
}

func getStock(s string, c chan string) {
        resp, err := http.Get("http://goanuj.freeshell.org/go/" + s + ".txt")
        if err != nil {
                log.Printf(s + ": " + err.Error())
                c <- err.Error() // channel send
                return
        }
        body, _ := ioutil.ReadAll(resp.Body)
        resp.Body.Close() // close ASAP to prevent too many open file desriptors
        val := string(body)
        //fmt.Printf("body: %s", val)
        c <- val // channel send
        return
}

func main() {
        start := time.Now()
        var sl = []string{"AAPL", "AMZN", "GOOG", "FB", "NFLX"}
        // creates slice of 1280 elements
        for i := 0; i < 8; i++ {
                sl = append(sl, sl...)
        }
        fmt.Printf("sl(size): %d\n", len(sl))

        // get channel that returns only strings
        c := getStocks(sl)
        for i := 0; i < len(sl); i++ {
                fmt.Printf("%s", <-c) // channel recv
        }

        fmt.Printf("main: %.2fs elapsed.\n", time.Since(start).Seconds())
}

Answer 1

不是为每个请求启动新的goroutine，而是在程序启动时创建一个固定池，并通过共享通道传递订单。每个订单都是一个与当前传递给getStock的参数相对应的结构。如果你需要能够杀死游泳池，情况会变得更加复杂，但它仍然不是那么难......

基本上你的新处理程序是循环，从所有处理程序共享的通道读取订单，执行它，然后在订单的响应通道上发送结果。

Answer 2

您需要使用缓冲通道来限制并行操作的数量。在开始循环中的新goroutine之前，您需要发送到此频道并在完成呼叫后从中接收，因此它将释放一个位置并且新请求将能够启动。查看您的修改后的代码on playground

golang：防止对等错误重置连接的策略

2 个答案: