如何在 Swift 中使用 substringToIndex?

本文介绍了如何在 Swift 中使用 substringToIndex?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我在这一行遇到编译器错误:

I get compiler error at this line:

UIDevice.currentDevice().identifierForVendor.UUIDString.substringToIndex(8)

类型String.Index"不符合协议IntegerLiteralConvertible"

我的目的是获取子字符串，但如何获取?

My intention is to get the substring, but how?

推荐答案

在 Swift 中，String 索引遵循字素簇，并且 IndexType 不是 Int.您有两种选择 - 将字符串(您的 UUID)转换为 NSString，并将其用作before"，或者创建第 n 个字符的索引.

In Swift, String indexing respects grapheme clusters, and an IndexType is not an Int. You have two choices - cast the string (your UUID) to an NSString, and use it as "before", or create an index to the nth character.

两者都如下图所示:

然而，该方法在 Swift 版本之间发生了根本性的变化.向下阅读以获取更高版本...

However, the method has changed radically between versions of Swift. Read down for later versions...

Swift 1

let s: String = "Stack Overflow" let ss1: String = (s as NSString).substringToIndex(5) // "Stack" //let ss2: String = s.substringToIndex(5)// 5 is not a String.Index let index: String.Index = advance(s.startIndex, 5) let ss2:String = s.substringToIndex(index) // "Stack"

CMD-单击 substringToIndex 会令人困惑地将您带到 NSString 定义，但是 CMD-单击 String 会发现以下内容:

CMD-Click on substringToIndex confusingly takes you to the NSString definition, however CMD-Click on String and you will find the following:

extension String : Collection { struct Index : BidirectionalIndex, Reflectable { func successor() -> String.Index func predecessor() -> String.Index func getMirror() -> Mirror } var startIndex: String.Index { get } var endIndex: String.Index { get } subscript (i: String.Index) -> Character { get } func generate() -> IndexingGenerator<String> }

斯威夫特 2正如评论员@DanielGalasko 指出的 advance 现在已经改变了......

Swift 2 As commentator @DanielGalasko points out advance has now changed...

let s: String = "Stack Overflow" let ss1: String = (s as NSString).substringToIndex(5) // "Stack" //let ss2: String = s.substringToIndex(5)// 5 is not a String.Index let index: String.Index = s.startIndex.advancedBy(5) // Swift 2 let ss2:String = s.substringToIndex(index) // "Stack"

斯威夫特 3在 Swift 3 中，它又发生了变化:

Swift 3 In Swift 3, it's changed again:

let s: String = "Stack Overflow" let ss1: String = (s as NSString).substring(to: 5) // "Stack" let index: String.Index = s.index(s.startIndex, offsetBy: 5) var ss2: String = s.substring(to: index) // "Stack"

斯威夫特 4在 Swift 4 中，还有另一个变化:

Swift 4 In Swift 4, yet another change:

let s: String = "Stack Overflow" let ss1: String = (s as NSString).substring(to: 5) // "Stack" let index: String.Index = s.index(s.startIndex, offsetBy: 5) var ss3: Substring = s[..<index] // "Stack" var ss4: String = String(s[..<index]) // "Stack"