在scala play框架异步操作中合并/链接未来

import scala.concurrent.Future
import scala.concurrent.ExecutionContext.Implicits.global

val future1: Future[Int] = Future.successful(1)
val future2: Future[Int] = Future.successful(2)
val future3: Future[Int] = Future.successful(3)

val mergedFuture: Future[Seq[Int]] = Future.sequence(Seq(future1, future2, future3))

mergedFuture.map { results =>
  // 处理合并后的结果
  println(results) // 输出: Seq(1, 2, 3)
}

import scala.concurrent.Future
import scala.concurrent.ExecutionContext.Implicits.global

val future1: Future[Int] = Future.successful(1)
val future2: Future[Int] = Future.successful(2)
val future3: Future[Int] = Future.successful(3)

val mergedFuture: Future[Int] = Future.fold(Seq(future1, future2, future3))(0) { (acc, value) =>
  // 对结果进行折叠操作
  acc + value
}

mergedFuture.map { result =>
  // 处理合并后的结果
  println(result) // 输出: 6
}

import scala.concurrent.Future
import scala.concurrent.ExecutionContext.Implicits.global

val future1: Future[Int] = Future.successful(1)
val future2: Future[Int] = Future.successful(2)

val mergedFuture: Future[(Int, Int)] = future1.zip(future2)

mergedFuture.map { results =>
  // 处理合并后的结果
  println(results) // 输出: (1, 2)
}