async Action api 如何在 play framework 2.2.x for scala 中工作?

How async Action api works in play framework 2.2.x for scala?

我正在尝试创建异步 api。但响应显示顺序执行。 完成的步骤: 在 chrome 的两个选项卡中打开 url。并迅速将他们一个接一个地击中。 url 例如:- localhost:9000/getStar.

但是执行日志是这样的:-

    [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

此代码是:

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也是顺序执行的?

Action.async 不会神奇地使控制器方法异步。 only 它的不同之处在于它需要 Future[Result] 而不是 Result。而已。控制器在其他方面是异步的,因为它们本质上是异步的(即一个普通的 Action 无论如何都会被包裹在 Future 中)。这里的问题是 Thread.sleep(1000) 阻塞 它是线程,一点也不异步。

另一件事是,在开发模式下(即activator run),播放服务器使用单线程来处理请求,因此它可以正确处理reload/compile、进化等。那是什么发生的是你只是用同步调用阻塞了那个线程。使用 activator start 应该会看到不同的结果,但即便如此,在这里使用 Action.async 也毫无意义,除非您要将该阻塞委托给不同的线程池。

Further reading.

只是为了澄清m-z的回答。 这是您如何在代码中处理一些异步集合的示例

def getStarAsyncOld(name: String) = Action.async {
  val futureResult = Future {
    Logger.info("Async started ************************** :" + name)
  } flatMap (_ => Future.sequence(for (a <- 1 until 10) yield Future {
    Thread.sleep(1000)
    Logger.info("Success Async call  :" + a.toString)
  })) map { _ =>
    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))
  }
}

或完全相同使用 for:

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

我做了很多实验,发现了一件事。也许这听起来很疯狂,但 Play 仅在 从同一浏览器到同一路由 时才按顺序处理并发请求。如果我通过 curl 或 从不同的浏览器 甚至从一个浏览器 向不同的路由 发出请求,然后他们被异步处理。不知道Play以这种方式做了什么样的保护,但这种保护存在并且是事实。