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。

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?

修改现有可变字符串的类别:

extension String
{
    mutating func replace(originalString:String, withString newString:String)
    {
        let replacedString = self.stringByReplacingOccurrencesOfString(originalString, withString: newString, options: nil, range: nil)
        self = replacedString
    }
}

使用:

name.replace(" ", withString: "+")

这在swift 4.2中很容易做到。只需使用replacingOccurrences(of: " "， with: "_")进行替换

var myStr = "This is my string"
let replaced = myStr.replacingOccurrences(of: " ", with: "_")
print(replaced)

斯威夫特5.5

我正在使用这个扩展:

extension String {

    func replaceCharacters(characters: String, toSeparator: String) -> String {
        let characterSet = CharacterSet(charactersIn: characters)
        let components = components(separatedBy: characterSet)
        let result = components.joined(separator: toSeparator)
        return result
    }

    func wipeCharacters(characters: String) -> String {
        return self.replaceCharacters(characters: characters, toSeparator: "")
    }
}

用法:

"<34353 43434>".replaceCharacters(characters: "< >", toSeparator:"+") // +34353+43434+
"<34353 43434>".wipeCharacters(characters: "< >") // 3435343434

我实现了这个非常简单的func:

func convap (text : String) -> String {
    return text.stringByReplacingOccurrencesOfString("'", withString: "''")
}

所以你可以这样写:

let sqlQuery = "INSERT INTO myTable (Field1, Field2) VALUES ('\(convap(value1))','\(convap(value2)')