scala：折叠元组成功

Question

使用返回Either [Fail，TupleX]的表达式，我如何折叠结果而不必在成功块中定义本地val？

// 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 
    ... 
    } 
) 

Answer 1

像这样

Answer 2

这里是一个最小的工作例如：

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