邏輯，如果裏面的比賽情況斯卡拉

Question

我需要用一大堆的比賽情況Scala中，我怎麼使用或如果匹配盒子內部，例如：

import scala.util.Random 

val x = Random.nextInt(30) 
val result = match x { 
case 0 to 10 => bad 
case 11 to 20 => average 
case 20 to 30 => cool 
} 

的第一個問題是怎麼做的我這樣做，而不是使用floor,ceil,round或其他數學的東西？

其次，對於字符串值，在PHP中，我們可以很容易地使用if (x == 'some')||(x == 'thing'){result}但這怎麼可以在scala匹配情況下工作？

例如當VAL x是隨機的A至I：

val result = match x { 
case A || B || C => bad 
case D || E || F => average 
case G || H || I => cool 
} 

謝謝！

Answer 1

這裏是一個更接近你的僞代碼的替代：

val x:Int 

val result = x match { 
    case x if (1 to 10).contains(x) => "bad" 
    case x if (11 to 20).contains(x) => "average" 
    case x if (21 to 30).contains(x) => "cool" 
    case _ => "unknown" 
} 


val y:String 

val result = y match { 
    case "A" | "B" | "C" => "bad" 
    case "D" | "E" | "F" => "average" 
    case "G" | "H" | "I" => "cool" 
    case _ => "unknown" 
} 

這將是更方便的治療字母分數作爲字符以利用範圍在第一個例子：

val z:Char 

val result = z match { 
    case x if ('A' to 'C').contains(x) => "bad" 
    case x if ('D' to 'F').contains(x) => "average" 
    case x if ('G' to 'I').contains(x) => "cool" 
    case _ => "unknown" 
} 

Answer 2

圖案使用逆天可以幫助你做到這

val x = Random.nextInt(30) 

val result = x match { 
case x if x > 0 && x <= 10 => "bad" 
case x if x > 11 && x <= 20 => "average" 
case x if x > 20 && x <= 30 => "cool" 
case _ => "no idea" 
} 

對於字符串，你可以做到這一點

val str = "foo" 

val result = str match { 
case "foo" => "found foo" 
case "bar" => "found bar" 
case _ => "found some thing" 
} 

您可以使用|匹配以及

val result = str match { 
    case "foo" | "bar" => "match" 
    case _ => "no match" 
} 

斯卡拉REPL

scala> :paste 
// Entering paste mode (ctrl-D to finish) 

val x = Random.nextInt(30) 

val result = x match { 
case x if x > 0 && x <= 10 => "bad" 
case x if x > 11 && x <= 20 => "average" 
case x if x > 20 && x <= 30 => "cool" 
case _ => "no idea" 
} 

// Exiting paste mode, now interpreting. 

x: Int = 13 
result: String = average