我有一个很长的计算算法,我想参与一个Play应用程序。
我想添加超时值,因此如果计算时间超过一段时间,则应该中断并显示一些错误消息。
查看Handling asynchronous results documentation - handling time-outs它解释了如何在长时间计算中创建超时。
但是,我注意到尽管用户收到超时消息,但计算不会中断,即日志消息会一直打印。
如何在超时提升后中断长计算?
示例控制器代码是:
object Application extends Controller {
def timeout(n:Integer) = Action.async {
val futureInt = scala.concurrent.Future { longComputation() }
val timeoutFuture = play.api.libs.concurrent.Promise.timeout("Oops", 1.second)
Future.firstCompletedOf(Seq(futureInt, timeoutFuture)).map {
case i: Int => Ok("Got result: " + i)
case t: String => InternalServerError(t)
}
}
def longComputation(): Int = {
while (true) {
Thread.sleep(1000)
Logger.debug("Computing...")
}
return 0
}
}
答案 0 :(得分:1)
为了满足此问题的要求,如果其运行时间超过最大持续时间,则必须能够中断长时间运行的计算。此外,有必要处理控制器动作中断的可能性。
假设计算涉及多个步骤和/或重复,一种中断此计算的方法(而不是仅仅放弃其结果并使计算保持运行)是定期检查计算的当前持续时间是否大于最大持续时间
为了明确表示此计算可能失败,可以将其声明为返回Try[T]
。
然后,操作可以在将来成功时检查计算尝试的结果,并为成功或失败的尝试生成适当的输出。
例如:
package controllers
import play.api._
import play.api.libs.concurrent.Akka
import play.api.libs.concurrent.Execution.Implicits.defaultContext
import play.api.mvc._
import play.api.Play.current
import scala.concurrent.duration._
import scala.concurrent.ExecutionContext
import scala.concurrent.Future
import scala.util._
object Application extends Controller {
def factorial(n: Int) = Action.async {
computeFactorial(n, 3.seconds).map { result =>
result match {
case Success(i) => Ok(s"$n! = $i")
case Failure(ex) => InternalServerError(ex.getMessage)
}
}
}
def computeFactorial(n: BigInt, timeout: Duration): Future[Try[BigInt]] = {
val startTime = System.nanoTime()
val maxTime = timeout.toNanos
def factorial(n: BigInt, result: BigInt = 1): BigInt = {
// Calculate elapsed time.
val elapsed = System.nanoTime() - startTime
Logger.debug(s"Computing factorial($n) with $elapsed nanoseconds elapsed.")
// Abort computation if timeout was exceeded.
if (elapsed > maxTime) {
Logger.debug(s"Timeout exceeded.")
throw new ComputationTimeoutException("The maximum time for the computation was exceeded.")
}
// Introduce an artificial delay so that less iterations are required to produce the error.
Thread.sleep(100)
// Compute step.
if (n == 0) result else factorial(n - 1, n * result)
}
Future {
try {
Success(factorial(n))
} catch {
case ex: Exception => Failure(ex)
}
}(Contexts.computationContext)
}
}
class ComputationTimeoutException(msg: String) extends RuntimeException(msg)
object Contexts {
implicit val computationContext: ExecutionContext = Akka.system.dispatchers.lookup("contexts.computationContext")
}
如果不要求将计算的结果明确标记为错误,并且Play的默认异步失败处理(返回500内部服务器错误)足够,则代码可以更简洁:
object Application extends Controller {
def factorial(n: Int) = Action.async {
computeFactorial(n, 3.seconds).map { i => Ok(s"$n! = $i") }
}
def computeFactorial(n: BigInt, timeout: Duration): Future[BigInt] = {
val startTime = System.nanoTime()
val maxTime = timeout.toNanos
def factorial(n: BigInt, result: BigInt = 1): BigInt = {
if (System.nanoTime() - startTime > maxTime) {
throw new RuntimeException("The maximum time for the computation was exceeded.")
}
Thread.sleep(100)
if (n == 0) result else factorial(n - 1, n * result)
}
Future { factorial(n) }(Akka.system.dispatchers.lookup("contexts.computationContext"))
}
}
示例在自定义上下文中运行计算,该自定义上下文提供的线程池与Play用于处理HTTP请求的线程池不同。有关详细信息,请参阅Understanding Play thread pools。上下文在application.conf
:
contexts {
computationContext {
fork-join-executor {
parallelism-factor=20
parallelism-max = 200
}
}
}
有关可下载的示例,请参阅this GitHub project。