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

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)
    }
}

Swift 4/5更短:

let string = "123456"
let firstThree = String(string.prefix(3)) //"123"
let lastThree = String(string.suffix(3)) //"456"

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

一些有用的扩展:

extension String {
    func substring(from: Int, to: Int) -> String {
        let start = index(startIndex, offsetBy: from)
        let end = index(start, offsetBy: to - from)
        return String(self[start ..< end])
    }

    func substring(range: NSRange) -> String {
        return substring(from: range.lowerBound, to: range.upperBound)
    }
}

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]