var str = "Hello, playground"
let indexcut = str.firstIndex(of: ",")
print(String(str[..<indexcut!]))
print(String(str[indexcut!...]))

你可以试着用这种方法，并会得到适当的结果。

快速4 5 5+

Last的子字符串

let str = "Hello World"
let removeFirstSix = String(str.dropFirst(6))
print(removeFirstSix) //World

来自First的子字符串

let removeLastSix = String(str.dropLast(6))
print(removeLastSix) //Hello

var str = "Hello, playground"
let indexcut = str.firstIndex(of: ",")
print(String(str[..<indexcut!]))
print(String(str[indexcut!...]))

你可以试着用这种方法，并会得到适当的结果。

你可以使用扩展类String来创建你的自定义subString方法，如下所示:

extension String {
    func subString(startIndex: Int, endIndex: Int) -> String {
        let end = (endIndex - self.count) + 1
        let indexStartOfText = self.index(self.startIndex, offsetBy: startIndex)
        let indexEndOfText = self.index(self.endIndex, offsetBy: end)
        let substring = self[indexStartOfText..<indexEndOfText]
        return String(substring)
    }
}

用这个方法你可以得到字符串的特定范围。你需要传递起始索引和你想要的字符总数。

extension String{
    func substring(fromIndex : Int,count : Int) -> String{
        let startIndex = self.index(self.startIndex, offsetBy: fromIndex)
        let endIndex = self.index(self.startIndex, offsetBy: fromIndex + count)
        let range = startIndex..<endIndex
        return String(self[range])
    }
}

substring(from: index)转换为[index…]

检查样品

let text = "1234567890"
let index = text.index(text.startIndex, offsetBy: 3)

text.substring(from: index) // "4567890"   [Swift 3]
String(text[index...])      // "4567890"   [Swift 4]