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

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

一个快速的解决方案沿着Sunkas的路线:

extension String {
    mutating func replace(_ originalString:String, with newString:String) {
        self = self.replacingOccurrences(of: originalString, with: newString)
    }
}

Use:

var string = "foo!"
string.replace("!", with: "?")
print(string)

输出:

foo?

斯威夫特4:

let abc = "Hello world"

let result = abc.replacingOccurrences(of: " ", with: "_", 
    options: NSString.CompareOptions.literal, range:nil)

print(result :\(result))

输出:

result : Hello_world

从Swift 2开始，String不再符合SequenceType。换句话说，你不能用for…在循环。

简单而简单的方法是将String转换为Array以获得索引的好处:

let input = Array(str)

我记得当我试图索引到字符串不使用任何转换。我真的很沮丧，因为我不能想出或达到一个理想的结果，我准备放弃了。 但我最终创建了我自己的解决方案，这里是扩展的完整代码:

extension String {
    subscript (_ index: Int) -> String {
    
        get {
             String(self[self.index(startIndex, offsetBy: index)])
        }
    
        set {
            remove(at: self.index(self.startIndex, offsetBy: index))
            insert(Character(newValue), at: self.index(self.startIndex, offsetBy: index))
        }
    }
}

现在，你可以像你最初想要的那样，使用索引从字符串中读取和替换单个字符:

var str = "cat"
for i in 0..<str.count {
 if str[i] == "c" {
   str[i] = "h"
 }
}

print(str)

这是一种简单而有用的方式来使用它，并通过Swift的字符串访问模型。 现在，下次你会觉得它很顺利，因为你可以循环字符串，而不是将它强制转换到数组中。

尝试一下，看看它是否有帮助!

这是一个在字符串上的替换方法的扩展，它没有不必要的复制，并在适当的地方做所有事情:

extension String {
    mutating func replaceOccurrences<Target: StringProtocol, Replacement: StringProtocol>(of target: Target, with replacement: Replacement, options: String.CompareOptions = [], locale: Locale? = nil) {
        var range: Range<Index>?
        repeat {
            range = self.range(of: target, options: options, range: range.map { self.index($0.lowerBound, offsetBy: replacement.count)..<self.endIndex }, locale: locale)
            if let range = range {
                self.replaceSubrange(range, with: replacement)
            }
        } while range != nil
    }
}

(方法签名也模仿内置String.replacingOccurrences()方法的签名)

可用于以下方式:

var string = "this is a string"
string.replaceOccurrences(of: " ", with: "_")
print(string) // "this_is_a_string"