Scala检查字符串是否包含特殊字符

Question

我编写了以下代码来检查字符串是否包含特殊字符.代码看起来太复杂了,但我不知道如何使它更简单.有任何想法吗？

def containsNoSpecialChars(string: String): Boolean = {
  val pattern = "^[a-zA-Z0-9]*$".r
  return pattern.findAllIn(string).mkString.length == string.length
}                                                 //> containsNoSpecialChars: (string: String)Boolean

containsNoSpecialChars("bl!a ")                   //> res0: Boolean = false
containsNoSpecialChars("bla9")                    //> res1: Boolean = true

Answer 1

这使用Java字符串:

word.matches("^[a-zA-Z0-9]*$")

或者如果您不想处理Regex,可以使用以下任一方法从Scala的RichString中受益:

word.forall(_.isLetterOrDigit)

要么:

!word.exists(!_.isLetterOrDigit)

Answer 2

scala> val ordinary=(('a' to 'z') ++ ('A' to 'Z') ++ ('0' to '9')).toSet
ordinary: scala.collection.immutable.Set[Char] = Set(E, e, X, s, x, 8, 4, n, 9, N, j, y, T, Y, t, J, u, U, f, F, A, a, 5, m, M, I, i, v, G, 6, 1, V, q, Q, L, b, g, B, l, P, p, 0, 2, C, H, c, W, h, 7, r, K, w, R, 3, k, O, D, Z, o, z, S, d)

scala> def isOrdinary(s:String)=s.forall(ordinary.contains(_))
isOrdinary: (s: String)Boolean


scala> isOrdinary("abc")
res4: Boolean = true

scala> isOrdinary("abc!")
res5: Boolean = false

我在逻辑上使用了Set,正确的选择,但它也应该使用Vector,这将避免你看到混乱的字母......

Answer 3

def containsNoSpecialChars(string: String) = string.matches("^[a-zA-Z0-9]*$")