从字符串中删除最后一个字符。 Swift语言 [英] Remove last character from string. Swift language

问题描述

如何使用Swift从String变量中删除最后一个字符？在文档中找不到。

以下是完整示例：

  var expression =45 + 22
 expression = expression.substringToIndex（countElements（expression） -  1）
  

解决方案

新的Swift 3.0方式

API已经获得更多 swifty ，因此Foundation扩展已更改了一些：

  var name：String = Dolphin
 var truncated = name.substring（to：name.index（before：name.endIndex））
 print（name）//Dolphin
 print（truncated） Dolphi
  

或就地版本：

  var name：String =Dolphin
 name.remove（at：name.index（before：name.endIndex））
 print（name） /Dolphi
  

感谢Zmey，Rob Allen！

Swift 2.0+ Way

有几种方法可以完成此操作：

通过Foundation扩展，尽管不是Swift库的一部分：

  var name：String =Dolphin
 var truncated = name.substringToIndex（name.endIndex.predecessor（））
 print（name）//Dolphin
 print（truncated） Dolphi
  

使用 removeRange（）方法（ 更改名称）：

  var name：String =Dolphin
 name.removeAtIndex（name.endIndex.predecessor（））
 print（name）//Dolphi
  / pre> 
 
 
使用 dropLast（）函数：
  var name：String =Dolphin
 var truncated = String（name.characters.dropLast（））
 print（name）//Dolphin
 print（truncated）//Dolphi
  
  Old String.Index 6 Beta 4 +）方式 
 
 
 
由于 String  Swift中的类型旨在提供出色的UTF-支持，则不能再使用 Int 类型访问字符索引/范围/子字符串。而是使用 String.Index ：
  let name：String = Dolphin
 let stringLength = count（name）//从swift1.2`countElements`成为`count` 
 let substringIndex = stringLength  -  1 
 name.substringToIndex ，substringIndex））//Dolphi
  
此外（更实用，您可以使用 endIndex ：
  let name：String =Dolphin 
 name.substringToIndex（name.endIndex.predecessor（））//Dolphi
  
 注意：我发现这是理解的理想起点， String.Index  
 
 
 
 旧版（预测试版4） 
 
 
 
你可以简单地使用 substringToIndex（）函数，它的长度小于 String ：
  let name：String =Dolphin
 name.substringToIndex（countElements（name） -  1） /Dolphi
  
 
How can I remove last character from String variable using Swift? Can't find it in documentation.

Here is full example:
var expression = "45+22"
expression = expression.substringToIndex(countElements(expression) - 1)

解决方案
New Swift 3.0 Way

The APIs have gotten a bit more swifty, and as a result the Foundation extension has changed a bit:
var name: String = "Dolphin"
var truncated = name.substring(to: name.index(before: name.endIndex))
print(name)      // "Dolphin"
print(truncated) // "Dolphi"
Or the in-place version:
var name: String = "Dolphin"
name.remove(at: name.index(before: name.endIndex))
print(name)      // "Dolphi"
Thanks Zmey, Rob Allen!

Swift 2.0+ Way

There are a few ways to accomplish this:

Via the Foundation extension, despite not being part of the Swift library:
var name: String = "Dolphin"
var truncated = name.substringToIndex(name.endIndex.predecessor())
print(name)      // "Dolphin"
print(truncated) // "Dolphi"
Using the removeRange() method (which alters the name):
var name: String = "Dolphin"    
name.removeAtIndex(name.endIndex.predecessor())
print(name) // "Dolphi"
Using the dropLast() function:
var name: String = "Dolphin"
var truncated = String(name.characters.dropLast())
print(name)      // "Dolphin"
print(truncated) // "Dolphi"
Old String.Index (Xcode 6 Beta 4 +) Way

Since String types in Swift aim to provide excellent UTF-8 support, you can no longer access character indexes/ranges/substrings using Int types. Instead, you use String.Index:
let name: String = "Dolphin"
let stringLength = count(name) // Since swift1.2 `countElements` became `count`
let substringIndex = stringLength - 1
name.substringToIndex(advance(name.startIndex, substringIndex)) // "Dolphi"
Alternatively (for a more practical, but less educational example) you can use endIndex:
let name: String = "Dolphin"
name.substringToIndex(name.endIndex.predecessor()) // "Dolphi"
Note: I found this to be a great starting point for understanding String.Index

Old (pre-Beta 4) Way

You can simply use the substringToIndex() function, providing it one less than the length of the String:
let name: String = "Dolphin"
name.substringToIndex(countElements(name) - 1) // "Dolphi"


                        
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