全部小写。小写()

首字母大写，其他字母小写，大写

全部大写。uppercase ()

Swift 3更新

replaceRange函数现在是replaceSubrange

nameOfString.replaceSubrange(nameOfString.startIndex...nameOfString.startIndex, with: String(nameOfString[nameOfString.startIndex]).capitalized)

Swift 3 (xcode 8.3.3)

大写字符串的所有首字符

let str = "your string"
let capStr = str.capitalized

//Your String

全部大写

let str = "your string"
let upStr = str.uppercased()

//YOUR STRING

字符串的第一个字符只能大写

 var str = "your string"
 let capStr = String(str.characters.prefix(1)).uppercased() + String(str.characters.dropFirst())

//Your string

下面是Swift 5的一个版本，如果第一个字母已经是大写，或者没有大小写的概念，使用Unicode标量属性API来保释:

extension String {
  func firstLetterUppercased() -> String {
    guard let first = first, first.isLowercase else { return self }
    return String(first).uppercased() + dropFirst()
  }
}

斯威夫特4.0

string.capitalized(with: nil)

or

string.capitalized

但是，每个单词的首字母都要大写

苹果公司的文档:

A capitalized string is a string with the first character in each word changed to its corresponding uppercase value, and all remaining characters set to their corresponding lowercase values. A “word” is any sequence of characters delimited by spaces, tabs, or line terminators. Some common word delimiting punctuation isn’t considered, so this property may not generally produce the desired results for multiword strings. See the getLineStart(_:end:contentsEnd:for:) method for additional information.

结合上面的答案，我写了一个小扩展，大写每个单词的第一个字母(因为这是我正在寻找的，并认为其他人可以使用它)。

我谦卑地提出:

extension String {
    var wordCaps:String {
        let listOfWords:[String] = self.componentsSeparatedByString(" ")
        var returnString: String = ""
        for word in listOfWords {
            if word != "" {
                var capWord = word.lowercaseString as String
                capWord.replaceRange(startIndex...startIndex, with: String(capWord[capWord.startIndex]).uppercaseString)
                returnString = returnString + capWord + " "
            }
        }
        if returnString.hasSuffix(" ") {
            returnString.removeAtIndex(returnString.endIndex.predecessor())
        }
        return returnString
    }
}