如何在Swift中找出字母是字母数字还是数字

问题描述

我要计算以下字符串中的字母，数字和特殊字符的数量：

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

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

我尝试过：

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

但是我遇到了错误。我对此尝试了各种其他变化-仍然出现错误-例如：

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

推荐答案

Swift 3的更新

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
    }
}

---

（以前的Swift版本的答案）

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++
    }
}

---

更新： 上述解决方案仅适用于ASCII字符集中的字符
，即不能将Ä，é或ø识别为字母。以下替代
解决方案使用Foundation框架中的 NSCharacterSet ，它可以根据字符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 alternativesolution uses NSCharacterSet from the Foundation framework, which can test charactersbased 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：从Xcode 6 beta 4开始，第一个解决方案不再起作用，因为
isAlpha （）和相关（仅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, becausethe isAlpha() and related (ASCII-only) methods have been removed from Swift.The second solution still works.

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