Swift：如何找出字母是字母数字还是数字 [英] Swift: how to find out if letter is Alphanumeric or Digit

问题描述

我想计算以下字符串中的字母，数字和特殊字符数：

I want to count the number of letters, digits and special characters in the following string:

let phrase = "The final score was 32-31!"

我试过：

for tempChar in phrase {
    if (tempChar >= "a" && tempChar <= "z") {
       letterCounter++
    }
// etc.

但我收到错误。我尝试了各种各样的其他变化 - 仍然得到错误 - 例如：

but I'm getting errors. I tried all sorts of other variations on this - still getting error - such as:

could not find an overload for '<=' that accepts the supplied arguments

任何线索？

推荐答案

一个可能的Swift解决方案：

A possible Swift solution:

var letterCounter = 0
var digitCount = 0
let phrase = "The final score was 32-31!"
for tempChar in phrase.unicodeScalars {
    if tempChar.isAlpha() {
        letterCounter++
    } else if tempChar.isDigit() {
        digitCount++
    }
}

---

上述解决方案仅适用于ASCII字符集中的字符
，即它不能将Ä，é或ø识别为字母。以下备选
解决方案使用Foundation框架中的 NSCharacterSet ，它可以根据其Unicode字符类测试字符
：

---

Update: The above solution works only with characters in the ASCII character set, i.e. it does not recognize Ä, é or ø as letters. The following alternative solution uses NSCharacterSet from the Foundation framework, which can test characters based on their Unicode character classes:

let letters = NSCharacterSet.letterCharacterSet()
let digits = NSCharacterSet.decimalDigitCharacterSet()

var letterCount = 0
var digitCount = 0

for uni in phrase.unicodeScalars {
    if letters.longCharacterIsMember(uni.value) {
        letterCount++
    } else if digits.longCharacterIsMember(uni.value) {
        digitCount++
    }
}

---

更新2：自Xcode 6 beta 4起，第一个解决方案不再工作，因为
isAlpha （）和相关（仅限ASCII）方法已从Swift中删除。
第二个解决方案仍然有效。

---

Update 2: As of Xcode 6 beta 4, the first solution does not work anymore, because the isAlpha() and related (ASCII-only) methods have been removed from Swift. The second solution still works.

Swift 3更新 p>

Update for Swift 3:

let letters = CharacterSet.letters
let digits = CharacterSet.decimalDigits

var letterCount = 0
var digitCount = 0

for uni in phrase.unicodeScalars {
    if letters.contains(uni) {
        letterCount += 1
    } else if digits.contains(uni) {
        digitCount += 1
    }
}

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