Scala 播放2.4 WebSocket演员不';不回答
我尝试将PlayFramework中的WebSocket与Akka Actor一起使用,但当我尝试将其与chrome或firefox一起使用时,它不起作用:Scala 播放2.4 WebSocket演员不';不回答,scala,playframework,websocket,akka,Scala,Playframework,Websocket,Akka,我尝试将PlayFramework中的WebSocket与Akka Actor一起使用,但当我尝试将其与chrome或firefox一起使用时,它不起作用: var exampleSocket = new WebSocket("ws://127.0.0.1:9000"); WebSocket connection to 'ws://127.0.0.1:9000/' failed: Error during WebSocket handshake: Unexpected response code
var exampleSocket = new WebSocket("ws://127.0.0.1:9000");
WebSocket connection to 'ws://127.0.0.1:9000/' failed: Error during WebSocket handshake: Unexpected response code: 200
我的控制器摘录:
package controllers
import javax.inject.Inject
import play.api.libs.ws.WSClient
import play.api.mvc.Controller
import akka.actor.ActorSystem
import actors.ContainersActor
import play.api.mvc.WebSocket
import play.api.Play.current
import play.api.libs.concurrent.Execution.Implicits.defaultContext
class ContainersController @Inject() (ws: WSClient, system: ActorSystem) extends Controller {
def socket = WebSocket.acceptWithActor[String, String] { request => out =>
ContainersActor.props(out)
}
println("socket : "+socket);
}
我的演员:
package actors
import akka.actor.Actor
import akka.actor.ActorRef
import akka.actor.Props
import akka.actor.actorRef2Scala
object ContainersActor {
def props(out: ActorRef) = Props(new ContainersActor(out))
}
class ContainersActor(out: ActorRef) extends Actor {
def receive = {
case msg: String =>
out ! ("I received your message: " + msg)
}
}
我遵循以下文件:
谢谢你的帮助:)
解决方案:
我没有为套接字创建路由,我以为它是另一个协议(x)
我在conf/routes中添加了这一行
GET /socket controllers.ContainersController.socket
我用以下命令调用套接字:
var exampleSocket = new WebSocket("ws://127.0.0.1:9000/socket");
您似乎正在尝试使用WebSocket连接连接到标准http路由,该路由会以状态代码200回答您。您能告诉我们您的路由吗?