2012-07-04 47 views
0

使用返回Either [Fail,TupleX]的表達式,我如何摺疊結果而不必在成功塊中定義本地val?scala:摺疊元組成功

// returns Either[Fail, Tuple2[String, String]] 
val result = for{ 
    model <- bindForm(form).right 
    key <- dao.storeKey(model.email, model.password) 
} yield (model.email, key) 

result fold (
    Conflict(_), 
    tuple2 => { // want to define email/key on this line 
    val(email,key) = tuple2 
    ... 
    } 
) 

回答

4

像這樣

+0

+1正是我一直在尋找,謝謝 – virtualeyes

2

這裏是一個最小的工作例如:

case class Conflict(s: String) 

def foo(result: Either[Conflict, Tuple2[String, String]]) = { 
    result.fold(
    c => println("left: " + c.toString), 
    { case (email, key) => println("right: %s, %s".format(email, key))} 
) 
} 

foo(Left(Conflict("Hi")))  // left: Conflict(Hi) 
foo(Right(("email", "key"))) // right: email, key 
+0

+1的「案(電子郵件,關鍵)=>「,已經休息了;-) – virtualeyes