Question

所以我试图把你的第一个字符串索引（fullName）并测试它以查看它是否匹配所有元音低和大写......出于某种原因，当我使用.startIndex我一次测试一个字母用if语句。

我有办法一次测试所有元音（）我试过||在每个之间它给了我错误＆＃34;无法将字符串转换为Bool＆＃34;。感谢任何帮助。

func lyricsForName(lyricsTemplate: String, fullName: String) -> String {

    let shortName = shortNameFromName(name: fullName)
    let index = fullName[fullName.startIndex]
    if index== ("a","A"){

        let lyrics = lyricsTemplate
            .replacingOccurrences(of:"<FULL_NAME>", with: fullName)
            .replacingOccurrences (of:"<SHORT_NAME>", with: fullName )

        return lyrics
    }else{
        let lyrics = lyricsTemplate
            .replacingOccurrences(of:"<FULL_NAME>", with: fullName)
            .replacingOccurrences (of:"<SHORT_NAME>", with: shortName )

        return lyrics

Answer 1

这是一种方法。您可以使用元音数组，然后测试数组是否包含第一个字符：

let fullName = "Albert"

let vowels = "aeiouAEIOU".characters    

if let first = fullName.characters.first, vowels.contains(first) {
    print("\(fullName) starts with a vowel")
}

Albert以元音开头

注意：使用fullName.characters.first比fullName[fullName.startIndex]更安全，因为后者会因空String而崩溃。

Answer 2

您可以像之前的答案那样做，但这是一种使用可用API的更快捷方式：

extension String {
    var firstThreeLettersAreVowels: Bool {
        guard characters.count >= 3 else {
            return false
        }

        let firstThreeLetters = substring(to: index(startIndex, offsetBy: 3))
        let isAllVowels = CharacterSet(charactersIn: firstThreeLetters).isSubset(of: CharacterSet.vowels)
        return isAllVowels
    }

    var isLower: Bool {
        return CharacterSet(charactersIn: self).isSubset(of: CharacterSet.lowercaseLetters)
    }

    var isUpper: Bool {
        return CharacterSet(charactersIn: self).isDisjoint(with: CharacterSet.lowercaseLetters)
    }
}

extension CharacterSet {
    static var vowels: CharacterSet {
        return CharacterSet(charactersIn: "AEIOUYaeiouy")
    }
}

或者一个班轮（没有长度检查）是

extension String {
    var firstThreeLettersAreVowels: Bool {
        return CharacterSet(charactersIn: substring(to: index(startIndex, offsetBy: 3)))
                    .isSubset(of: CharacterSet(charactersIn: "AEIOUYaeiouy"))
    }
}

这允许您依赖现有的API，而不是自己尝试工作，而且应该非常快。

Answer 3

所有荣誉请到PEEJWEEJ回答

字符集是他的想法：）

我只是更通用了

import Foundation

extension String {
    func isFirstLetters(count: Int, of characterSet: CharacterSet) -> Bool {

        guard characters.count >= count else {
            return false
        }

        let firstLetters = substring(to: index(startIndex, offsetBy: count))
        let isInSubset = CharacterSet(charactersIn: firstLetters).isSubset(of: characterSet

        )
        return isInSubset
    }
}

extension CharacterSet {
    static var vowels: CharacterSet {
        return CharacterSet(charactersIn: "AEIOUYaeiouy")
    }
}

一些测试用例：

"abcDEf".isFirstLetters(count: 3, of: .vowels) // false
"aioDEf".isFirstLetters(count: 3, of: .vowels) // true

"abcDEf".isFirstLetters(count: 1, of: .vowels) // true
"bbcDEf".isFirstLetters(count: 1, of: .vowels) // false
"ibcDEf".isFirstLetters(count: 1, of: .vowels) // true

"ABcdef".isFirstLetters(count: 3, of: .uppercaseLetters) // false
"ABDdef".isFirstLetters(count: 3, of: .uppercaseLetters) // true

"abBABcdef".isFirstLetters(count: 3, of: .lowercaseLetters) // false
"abbABDdef".isFirstLetters(count: 3, of: .lowercaseLetters) // true


"ABD".isFirstLetters(count: 3, of: CharacterSet(charactersIn: "A"..."C")) // false
"ABD".isFirstLetters(count: 3, of: CharacterSet(charactersIn: "A"..."D")) // true

调试信息：

extension CharacterSet {
    var characters: [Character] {
        var result: [Character] = []
        for plane: UInt8 in 0...16 where self.hasMember(inPlane: plane) {
            for unicode in UInt32(plane) << 16 ..< UInt32(plane + 1) << 16 {
                if let uniChar = UnicodeScalar(unicode), self.contains(uniChar) {
                    result.append(Character(uniChar))
                }
            }
        }
        return result
    }
}



print(CharacterSet.uppercaseLetters.description)
//<Foundation._SwiftNSCharacterSet: 0x6000000271a0>

print(String(CharacterSet.vowels.characters))
//AEIOUYaeiouy

print(CharacterSet.vowels.characters)
// ["A", "E", "I", "O", "U", "Y", "a", "e", "i", "o", "u", "y"]

将字符串转换为Bool并在Swift

3 个答案: