Question

我们正在使用gin在生产中公开一些REST API。现在，一旦HTTP服务器启动，我就必须做一些事情。

我对渠道不是很熟悉，但是下面给出的代码是我想要做的事情。 startHTPPRouter()启动HTTP服务后，我想向main()发送信号。根据该信号，我想做其他一些事情。

请让我知道下面给出的代码在做什么错。

func startHTTPRouter(routerChannel chan bool){
    router := gin.New()
    // Many REST API routes definitions
    router.Run("<port>")
    routerChannel <- true  // Is this gonna work ? Because Run() again launches a go routine for Serve()
}

func main() {
    routerChannel := make(chan bool)
    defer close(routerChannel)
    go startHTTPRouter(routerChannel )
    for {
        select {
        case <-routerChannel:
            doStuff()  // Only when the REST APIs are available.
            time.Sleep(time.Second * 5)
        default:
            log.Info("Waiting for router channel...")
            time.Sleep(time.Second * 5)
        }
    }
}

Answer 1

gin.New（）。Run（）阻止了API。杜松子酒服务器直到退出都不会返回。

func startHTTPRouter(routerChannel chan bool) {
    router := gin.New()
    router.Run("<port>")
    routerChannel <- true  // Is this gonna work ? Because Run() again launches a go routine for Serve()
}

下面是gin'Run（）API。 https://github.com/gin-gonic/gin/blob/master/gin.go

// Run attaches the router to a http.Server and starts listening and serving HTTP requests.
// It is a shortcut for http.ListenAndServe(addr, router)
// Note: this method will block the calling goroutine indefinitely unless an error happens.
func (engine *Engine) Run(addr ...string) (err error) {
    defer func() { debugPrintError(err) }()

    address := resolveAddress(addr)
    debugPrint("Listening and serving HTTP on %s\n", address)
    err = http.ListenAndServe(address, engine)
    return
}

等待gin HTTP服务器启动

1 个答案: