我有以下简单的代码写在Swift 3:

let str = "Hello, playground"
let index = str.index(of: ",")!
let newStr = str.substring(to: index)

在Xcode 9 beta 5中,我得到了以下警告:

'substring(to:)'已弃用:请使用带有'partial range from'操作符的字符串切片下标。

这个部分范围的切片下标如何在Swift 4中使用?


当前回答

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)
    }
}

其他回答

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])
    }
}

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)
    }
}

转换子字符串(Swift 3)到字符串切片(Swift 4)

在Swift 3,4中:

let newStr = str.substring(to: index) // Swift 3
let newStr = String(str[..<index]) // Swift 4

let newStr = str.substring(from: index) // Swift 3
let newStr = String(str[index...]) // Swift 4 

let range = firstIndex..<secondIndex // If you have a range
let newStr = = str.substring(with: range) // Swift 3
let newStr = String(str[range])  // Swift 4

如果你只是想获取一个特定字符的子字符串,你不需要先找到索引,你可以只使用prefix(while:)方法

let str = "Hello, playground"
let subString = str.prefix { $0 != "," } // "Hello" as a String.SubSequence

我使用的简单方法是:

String(Array(str)[2...4])