async Action api如何在scala的play framework 2.2.x中运行？

Question

我试图创建异步api.但响应显示顺序执行.完成的步骤:在两个chrome选项卡中打开url.然后快速地打击他们.url ex: - localhost:9000/getStar.

但执行日志如下: -

    [info] play - Listening for HTTP on /0:0:0:0:0:0:0:0:9000

(Server started, use Ctrl+D to stop and go back to the console...)

[success] Compiled in 107ms
[info] application - Application has started
[info] play - Application started (Dev)
[info] application - Async started ************************** :tarun
[info] application - Success Async call  :1
[info] application - Success Async call  :2
[info] application - Success Async call  :3
[info] application - Success Async call  :4
[info] application - Success Async call  :5
[info] application - Success Async call  :6
[info] application - Success Async call  :7
[info] application - Success Async call  :8
[info] application - Success Async call  :9
[info] application - Async finished ************************** :tarun
[info] application - Async started ************************** :tarun1
[info] application - Success Async call  :1
[info] application - Success Async call  :2
[info] application - Success Async call  :3
[info] application - Success Async call  :4
[info] application - Success Async call  :5
[info] application - Success Async call  :6
[info] application - Success Async call  :7
[info] application - Success Async call  :8
[info] application - Success Async call  :9
[info] application - Async finished ************************** :tarun1

这个代码是:

package controllers

import play.Logger
import play.api.libs.json.Json
import play.api.mvc._


import scala.concurrent.Future

object StarController extends Controller {
  import play.api.libs.concurrent.Execution.Implicits.defaultContext


  def getStarAsync(name : String) = Action.async{
    val futureResult = Future{
      Logger.info("Async started ************************** :" + name)
      val a = 0;
      for( a <- 1 until 10) {
        Thread.sleep(1000)
        Logger.info("Success Async call  :" + a.toString)
      }
      Logger.info("Async finished ************************** :" + name)
      Map("success" -> Json.toJson(true), "msg" -> Json.toJson("Success Async by :" + name), "code" -> Json.toJson(200))
    }

    futureResult.map{ result =>
      Ok(Json.toJson(result))
    }
  }

}

任何人都可以帮助我理解,为什么即使使用异步调用执行顺序？

Answer 1

Action.async不会神奇地使控制器方法异步.它唯一不同的是它期望一个Future[Result]而不是一个Result.而已.控制器在其他方面是异步的,因为它们本质上是可能的(即正常情况下Action会被包裹起来Future).这里的事情是Thread.sleep(1000) 阻塞它的线程,并且不是异步的.

另一件事是在开发模式(即activator run)中,播放服务器使用单个线程来处理请求,因此它可以正确处理重载/编译,演进等.所以正在发生的事情是你只是用同步来阻止该线程调用.你应该看到使用不同的结果activator start,但即便如此,Action.async除非你要将阻塞委托给不同的线程池,否则在这里使用是没有意义的.

进一步 阅读.