在 Scala Play 2.2.x 测试的背景下，我有一个Map[String, String]我需要将其传递给接受的函数(String, String)*即一个可变参数(String, String) tuple.
e.g.
val data : Map[String, String] = Map("value" -> "25", "id" -> "", "columnName" -> "trades")
route(FakeRequest(POST, "/whatever/do").withFormUrlEncodedBody(data))
但这会导致类型不匹配，因为withFormUrlEncodedBody只接受一个(String, String)* type.
Simply:
def foo(names: (String, String)*) = names.foreach(println)
val folks = Map("john" -> "smith", "queen" -> "mary")
foo(folks.toSeq:_*)
// (john,smith)
// (queen,mary)
Where _* 是对编译器的提示 https://stackoverflow.com/questions/6051302/what-does-colon-underscore-star-do-in-scala.
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