如果你不想使用Objective-C NSString方法，你可以使用split和join:

var string = "This is my string"
string = join("+", split(string, isSeparator: { $0 == " " }))

split(string, isSeparator:{$0 == " "})返回一个字符串数组(["This"， "is"， "my"， "string"])。

join用一个+连接这些元素，产生所需的输出:"This+is+my+string"。

你测试过这个吗?

var test = "This is my string"

let replaced = test.stringByReplacingOccurrencesOfString(" ", withString: "+", options: nil, range: nil)

你可以测试这个:

let newString = test。stringByReplacingOccurrencesOfString(" "， withString: "+"，选项:nil，范围:nil)

基于Ramis回答的Swift 3解决方案:

extension String {
    func withReplacedCharacters(_ characters: String, by separator: String) -> String {
        let characterSet = CharacterSet(charactersIn: characters)
        return components(separatedBy: characterSet).joined(separator: separator)
    }
}

尝试根据Swift 3的命名约定，提出一个合适的函数名。

Xcode 11•Swift 5.1

StringProtocol replacingOccurrences的突变方法可以实现如下:

extension RangeReplaceableCollection where Self: StringProtocol {
    mutating func replaceOccurrences<Target: StringProtocol, Replacement: StringProtocol>(of target: Target, with replacement: Replacement, options: String.CompareOptions = [], range searchRange: Range<String.Index>? = nil) {
        self = .init(replacingOccurrences(of: target, with: replacement, options: options, range: searchRange))
    }
}

var name = "This is my string"
name.replaceOccurrences(of: " ", with: "+")
print(name) // "This+is+my+string\n"

很少发生在我身上，我只是想改变字符串中的(一个单词或字符)

所以我用了字典

  extension String{
    func replace(_ dictionary: [String: String]) -> String{
          var result = String()
          var i = -1
          for (of , with): (String, String)in dictionary{
              i += 1
              if i<1{
                  result = self.replacingOccurrences(of: of, with: with)
              }else{
                  result = result.replacingOccurrences(of: of, with: with)
              }
          }
        return result
     }
    }

使用

let mobile = "+1 (800) 444-9999"
let dictionary = ["+": "00", " ": "", "(": "", ")": "", "-": ""]
let mobileResult = mobile.replace(dictionary)
print(mobileResult) // 001800444999