火了，忘了戈罗丁戈朗

Question

我编写了一个 API，用于进行数据库调用并执行一些业务逻辑。我正在调用一个必须在后台执行某些操作的戈鲁廷。由于 API 调用不应等待此后台任务完成，因此我在调用 goroutine 后立即返回 200 OK（让我们假设后台任务永远不会给出任何错误）。

我读到戈鲁廷将在戈鲁廷完成其任务后被终止。这种火灾和忘记远离戈鲁丁泄漏是安全的吗？一旦戈鲁廷执行了工作，它们是否会被终止和清理？

func DefaultHandler(w http.ResponseWriter, r *http.Request) {

    // Some DB calls

    // Some business logics

    go func() {

        // some Task taking 5 sec

    }()

    w.WriteHeader(http.StatusOK)

}

Answer 1

我建议始终控制您的戈鲁丁，以避免内存和系统耗尽。如果您收到请求激增，并且开始不受控制地生成goroutines，则系统可能很快就会关闭。在那些需要立即返回200Ok的情况下，最好的方法是创建一个消息队列，因此服务器只需要在队列中创建一个作业并返回ok和忘记。其余部分将由使用者异步处理。创建者（HTTP 服务器）>>>队列>>>使用者通常，队列是外部资源 （RabbitMQ， AWS SQS...），但出于教学目的，您可以使用与消息队列相同的通道实现相同的效果。在示例中，您将看到我们如何创建一个通道来传达 2 个进程。然后，我们启动将从通道读取的工作进程，然后使用将写入通道的处理程序启动服务器。尝试在发送 curl 请求时使用缓冲区大小和作业时间。package mainimport (&nbsp; &nbsp; "fmt"&nbsp; &nbsp; "log"&nbsp; &nbsp; "net/http"&nbsp; &nbsp; "time")/*$ go run .curl "http://localhost:8080?user_id=1"curl "http://localhost:8080?user_id=2"curl "http://localhost:8080?user_id=3"curl "http://localhost:8080?user_id=....."*/func main() {&nbsp; &nbsp; queueSize := 10&nbsp; &nbsp; // This is our queue, a channel to communicate processes. Queue size is the number of items that can be stored in the channel&nbsp; &nbsp; myJobQueue := make(chan string, queueSize) // Search for 'buffered channels'&nbsp; &nbsp; // Starts a worker that will read continuously from our queue&nbsp; &nbsp; go myBackgroundWorker(myJobQueue)&nbsp; &nbsp; // We start our server with a handler that is receiving the queue to write to it&nbsp; &nbsp; if err := http.ListenAndServe("localhost:8080", myAsyncHandler(myJobQueue)); err != nil {&nbsp; &nbsp; &nbsp; &nbsp; panic(err)&nbsp; &nbsp; }}func myAsyncHandler(myJobQueue chan<- string) http.HandlerFunc {&nbsp; &nbsp; return func(rw http.ResponseWriter, r *http.Request) {&nbsp; &nbsp; &nbsp; &nbsp; // We check that in the query string we have a 'user_id' query param&nbsp; &nbsp; &nbsp; &nbsp; if userID := r.URL.Query().Get("user_id"); userID != "" {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; select {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; case myJobQueue <- userID: // We try to put the item into the queue ...&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; rw.WriteHeader(http.StatusOK)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; rw.Write([]byte(fmt.Sprintf("queuing user process: %s", userID)))&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; default: // If we cannot write to the queue it's because is full!&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; rw.WriteHeader(http.StatusInternalServerError)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; rw.Write([]byte(`our internal queue is full, try it later`))&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; rw.WriteHeader(http.StatusBadRequest)&nbsp; &nbsp; &nbsp; &nbsp; rw.Write([]byte(`missing 'user_id' in query params`))&nbsp; &nbsp; }}func myBackgroundWorker(myJobQueue <-chan string) {&nbsp; &nbsp; const (&nbsp; &nbsp; &nbsp; &nbsp; jobDuration = 10 * time.Second // simulation of a heavy background process&nbsp; &nbsp; )&nbsp; &nbsp; // We continuosly read from our queue and process the queue 1 by 1.&nbsp; &nbsp; // In this loop we could spawn more goroutines in a controlled way to paralelize work and increase the read throughput, but i don't want to overcomplicate the example.&nbsp; &nbsp; for userID := range myJobQueue {&nbsp; &nbsp; &nbsp; &nbsp; // rate limiter here ...&nbsp; &nbsp; &nbsp; &nbsp; // go func(u string){&nbsp; &nbsp; &nbsp; &nbsp; log.Printf("processing user: %s, started", userID)&nbsp; &nbsp; &nbsp; &nbsp; time.Sleep(jobDuration)&nbsp; &nbsp; &nbsp; &nbsp; log.Printf("processing user: %s, finisehd", userID)&nbsp; &nbsp; &nbsp; &nbsp; // }(userID)&nbsp; &nbsp; }}

Answer 2

您无需处理“戈鲁丁清理”，您只需启动戈鲁丁，当作为戈鲁廷返回的功能启动时，它们将被清除。引用规范：围棋声明：当函数终止时，其戈鲁廷也终止。如果函数具有任何返回值，则在函数完成时将丢弃这些返回值。所以你所做的很好。但请注意，您启动的 goroutine 不能使用或假定有关请求 （） 和响应编写器 （） 的任何内容，您只能在从处理程序返回之前使用它们。rw另请注意，您不必编写 ，如果您从处理程序返回而不写入任何内容，则假定为成功并将自动发送回去。http.StatusOKHTTP 200 OK查看相关/可能的重复项：在另一个戈鲁廷上运行的 Webhook 进程

Answer 3

没有“戈鲁丁清洁”，你可以使用网络钩子或像gocraft这样的后台工作。我能想到的使用你的解决方案的唯一方法是将同步包用于学习目的。func DefaultHandler(w http.ResponseWriter, r *http.Request) {// Some DB calls// Some business logicsvar wg sync.WaitGroupwg.Add(1)go func() {&nbsp; defer wg.Done()&nbsp; &nbsp; // some Task taking 5 sec}()w.WriteHeader(http.StatusOK)wg.wait()}

Answer 4

您可以等待戈鲁廷完成使用：&sync.WaitGroup// BusyTaskfunc BusyTask(t interface{}) error {&nbsp; &nbsp; var wg = &sync.WaitGroup{}&nbsp; &nbsp; wg.Add(1)&nbsp; &nbsp; go func() {&nbsp; &nbsp; &nbsp; &nbsp; // busy doing stuff&nbsp; &nbsp; &nbsp; &nbsp; time.Sleep(5 * time.Second)&nbsp; &nbsp; &nbsp; &nbsp; wg.Done()&nbsp; &nbsp; }()&nbsp; &nbsp; wg.Wait() // wait for goroutine&nbsp; &nbsp; return nil}// this will wait 5 second till goroutune finishfunc main() {&nbsp; &nbsp; fmt.Println("hello")&nbsp; &nbsp; BusyTask("some task...")&nbsp; &nbsp; fmt.Println("done")}另一种方法是将 a 附加到戈鲁丁并超时。context.Context//func BusyTaskContext(ctx context.Context, t string) error {&nbsp; &nbsp; done := make(chan struct{}, 1)&nbsp; &nbsp; //&nbsp; &nbsp; go func() {&nbsp; &nbsp; &nbsp; &nbsp; // time sleep 5 second&nbsp; &nbsp; &nbsp; &nbsp; time.Sleep(5 * time.Second)&nbsp; &nbsp; &nbsp; &nbsp; // do tasks and signle done&nbsp; &nbsp; &nbsp; &nbsp; done <- struct{}{}&nbsp; &nbsp; &nbsp; &nbsp; close(done)&nbsp; &nbsp; }()&nbsp; &nbsp; //&nbsp; &nbsp; select {&nbsp; &nbsp; case <-ctx.Done():&nbsp; &nbsp; &nbsp; &nbsp; return errors.New("timeout")&nbsp; &nbsp; case <-done:&nbsp; &nbsp; &nbsp; &nbsp; return nil&nbsp; &nbsp; }}//func main() {&nbsp; &nbsp; fmt.Println("hello")&nbsp; &nbsp; ctx, cancel := context.WithTimeout(context.TODO(), 2*time.Second)&nbsp; &nbsp; defer cancel()&nbsp; &nbsp; if err := BusyTaskContext(ctx, "some task..."); err != nil {&nbsp; &nbsp; &nbsp; &nbsp; fmt.Println(err)&nbsp; &nbsp; &nbsp; &nbsp; return&nbsp; &nbsp; }&nbsp; &nbsp; fmt.Println("done")}