scala - 如何克服scala中的类型不匹配

Question

def getDesiredValue(measure: String, value: String): Any = {

    val desiredMeasure = List("temperature","speed","length")
    if(desiredMeasure.contains(measure)){
        measure match
        {
            case temperature => value
            case speed => value
            case length => value
        }
        value.toString
    }

}

val Value = getDesiredValue("temperature", "°C")

编译错误：

found : Unit
required: String

if(desiredMeasure.contains(measure)){
^

虽然“值”被传递给函数，但它应该只返回列表中的元素（“温度”、“速度”、“长度”），绕过其余的输入。

标签： scala

Answer 1

这是构建此代码的更好方法，并且应该修复您看到的错误：

def getDesiredValue(measure: String, value: String) =
  measure match {
    case "temperature" => value
    case "speed" => value
    case "length" => value
    case _ => errorValue
  }