我正在寻找一种方法来取代字符在一个Swift字符串。

示例:“This is my string”

我想用“+”替换“”以获得“This+is+my+string”。

我怎样才能做到这一点呢?


当前回答

你可以测试这个:

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

其他回答

迅速扩展:

extension String {

    func stringByReplacing(replaceStrings set: [String], with: String) -> String {
        var stringObject = self
        for string in set {
            stringObject = self.stringByReplacingOccurrencesOfString(string, withString: with)
        }
        return stringObject
    }

}

继续使用let replacedString = yorString。stringbyreplacement (replaceStrings: [" ","?","."],加上:“+”)

函数的速度是我几乎不能感到骄傲的,但是你可以在一次传递一个String数组来进行多次替换。

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)

你可以用这个:

let s = "This is my string"
let modified = s.replace(" ", withString:"+")    

如果你在你的代码中添加这个扩展方法:

extension String
{
    func replace(target: String, withString: String) -> String
    {
       return self.stringByReplacingOccurrencesOfString(target, withString: withString, options: NSStringCompareOptions.LiteralSearch, range: nil)
    }
}

斯威夫特3:

extension String
{
    func replace(target: String, withString: String) -> String
    {
        return self.replacingOccurrences(of: target, with: withString, options: NSString.CompareOptions.literal, range: 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。