Playframework:发现类型不匹配scala.concurrent.Future[play.api.mvc.Result]必需:play.api.mvc.Result
我在PlayFramework中的控制器中有以下代码:Playframework:发现类型不匹配scala.concurrent.Future[play.api.mvc.Result]必需:play.api.mvc.Result,scala,playframework-2.5,Scala,Playframework 2.5,我在PlayFramework中的控制器中有以下代码: def auth = Action.async(parse.json) { request => { val authRequest = request.body.validate[AuthRequest] authRequest.fold( errors => Future(BadRequest), auth => { creden
def auth = Action.async(parse.json) { request =>
{
val authRequest = request.body.validate[AuthRequest]
authRequest.fold(
errors => Future(BadRequest),
auth => {
credentialsManager.checkEmailPassword(auth.email, auth.password).map {
case Some(credential: Credentials) => {
sessionManager.createSession(credential.authAccountId).map { //Throws an error
case Some(authResponse: AuthResponse) => Ok(Json.toJson(authResponse))
case None => InternalServerError
}
}
case (None) => Unauthorized
}
})
}
}
我在上面带有错误注释的行中得到以下错误:
Type Mismatch:
[error] found : scala.concurrent.Future[play.api.mvc.Result]
[error] required: play.api.mvc.Result
[error] sessionManager.createSession(credential.authAccountId).map {
那里的createSession调用返回一个Future[Option[Object]]
,但我不知道如何解决这个问题
非常感谢您的帮助。不确定,但这应该可以:
def auth = Action.async(parse.json) { request =>
{
val authRequest = request.body.validate[AuthRequest]
authRequest.fold(
errors => Future(BadRequest),
auth => {
credentialsManager.checkEmailPassword(auth.email, auth.password).flatMap { //flatMap
case Some(credential: Credentials) => {
sessionManager.createSession(credential.authAccountId).map {
case Some(authResponse: AuthResponse) => Ok(Json.toJson(authResponse))
case None => InternalServerError
}
}
case None => Future(Unauthorized) //Wrap it
}
})
}
}
这是对代码的简化,并附有一些注释。我希望这足以抓住这个想法:
Future(Option("validCredentials")).flatMap {
case Some(credential) => Future("OK")
case None => Future("Unauthorized")
}
//Future[Option[String]].flatMap(Option[String] => Future[String])
//Future[A].flatMap(A => Future[B]) //where A =:= Option[String] and B =:= String
简短答复:
将凭证管理器中的.map
更改为.flatMap
。检查电子邮件密码(auth.email,auth.password)。将和案例(无)=>未经授权的更改为案例无=>未来(未经授权的)
说明:
credentialsManager.checkEmailPassword(auth.email,auth.password)
返回一个Future[Option[Credentials]]
映射,该映射将始终返回一个Future
并在其内部sessionManager.createSession(credential.authcountId)
还返回一个Future
So,credentialmanager.checkEmailPassword(auth.email,auth.password)
的最终结果是Future[Future[something]
为了避免这种情况,你可以将它展平,然后将它映射到map
Future(“未经授权”)一步完成无效。请使用Future.successful(确定(“确定”))
Future.successful(BadRequest(Unauthorized))您的credentialsManager.checkEmailPassword
方法返回什么?它返回Future[Option[Credentials]]
并定义为def checkEmailPassword(电子邮件:字符串,密码:字符串):Future[Option[Credentials]]
将.map
更改为.flatMap
中的credentialsManager。检查电子邮件密码(auth.email,auth.password)。将和案例(无)=>未经授权的转换为案例无=>未来(未经授权)
真棒!成功了。你能解释一下背后的原因吗?我假设它会将外部的Future[选项[凭证]]
展平,但这会如何影响内部映射/返回类型?值得注意的是,Scala 2.12中将提供期货的展平。现在您必须flatMap(identity)
好了,map
,flatMap
和filter
是monad本身可以使用的一些操作。谢谢您的解释!