我一直在尝试创建一个需要尽快扫描网络(主要是LAN)上的开放端口的应用程序。
我搜索过,我找到的一个很好的方法使用以下代码:
(1 to 65536).par.map { case port ⇒
try {
val socket = new java.net.Socket("127.0.0.1", port)
socket.close()
println(port)
port
} catch {
case _: Throwable ⇒ -1
}
}.toSet
但是,代码的问题是,如果我输入127.0.0.1以外的任何内容或localhost作为位置(比如192.168.1.2),应用程序会冻结。
知道为什么会发生这种情况以及如何解决这个问题?
P.S。我也尝试用socket.setSoTimeout(1500)设置套接字超时,但没有改变。
答案 0 :(得分:6)
像
这样的东西import scala.concurrent.{Future, Await}
import scala.concurrent.duration._
import scala.util.Try
import scala.concurrent._
import java.util.concurrent.Executors
implicit val ec = ExecutionContext.fromExecutor(Executors.newFixedThreadPool(100))
def openPorts(address:String ="127.0.0.1",duration:Duration = 10 seconds, fromPort:Int = 1, toPort:Int = 65536) = {
val socketTimeout = 200
val result = Future.traverse(fromPort to toPort ) { port =>
Future{ Try {
val socket = new java.net.Socket()
socket.connect(new java.net.InetSocketAddress(address, port),socketTimeout)
socket.close()
port
} toOption }
}
Try {Await.result(result, duration)}.toOption.getOrElse(Nil).flatten
}
scala> val localPorts openPorts(fromPort = 10, toPort = 1000)
localPorts: scala.collection.immutable.IndexedSeq[Int] = Vector(22, 631)
scala> val remotePorts = openPorts(fromPort = 10, toPort = 1000, address="192.168.1.20")
remotePorts: scala.collection.immutable.Seq[Int] = List() //we ate the timeout
scala> val remotePorts = openPorts(fromPort = 12000, toPort = 13000, address="91.190.218.61", duration=30 seconds)
remotePorts: scala.collection.immutable.Seq[Int] = Vector(12345, 12350)
答案 1 :(得分:2)
虽然Ashalynd的答案也很有效,但当我再次使用这个应用程序时,我找到了一个解决方案,或许更简单,使用Futures。
这是代码
import scala.concurrent._
import ExecutionContext.Implicits.global
object TestOne extends App{
println("Program is now running")
val dataset: Future[Set[Int]] = future {
(1 to 6335).map {
case port =>
try {
val socket = new java.net.Socket("127.0.0.1", port)
socket.close()
println("The port is open at " + port)
port
} catch {
case _: Throwable => -1
}
}.toSet
}
}