如何使用 if guard 进行推导?
type Error = String
type Success = String
def csrfValidation(session:Session, body:JsValue):Either[Error,Success] = {
val csrfRet = for (csrfSession <- csrfStateSessionValidation(session).right;
csrfReq <- csrfStateReqBodyValidation(body).right if (csrfSession == csrfReq)) yield (csrfReq)
if (csrfRet.isRight)
Right(csrfRet)
else {
Logger.warn("request and session csrf is not the same")
Left("Oops,something went wrong, request and session csrf is not the same")
}
}
我在使用它时遇到了这个错误。
'withFilter' method does not yet exist on scala.util.Either.RightProjection[Error,Success], using `filter' method instead
编辑: 我遇到了另一个错误。我认为如果使用防护,它会返回一个选项结果。
[error] type mismatch;
[error] found : Option[scala.util.Either[Nothing,controllers.ProfileApiV1.Success]]
[error] required: scala.util.Either[?,?]
[error] csrfReq <- csrfStateReqBodyValidation(body).right if (csrfSession == csrfReq)) yield (csrfReq)
编辑2
This is what I did to fix above error. I also move if-guard to later process.
val result = for {
foo <- Right[String,String]("teststring").right
bar <- Right[String,String]("teststring").right
} yield (foo, bar)
result fold (
ex => Left("Operation failed with " + ex),
v => v match {
case (x,y) =>
if (x == y) Right(x)
else Left("value is different")
}
)