删除Swift 3中的最后一个字符 [英] Delete last character in Swift 3

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本文介绍了删除Swift 3中的最后一个字符的处理方法,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

我正在创建一个简单的计算器应用程序,目前正在努力删除我的按钮被点击时的最后一个字符。我正在使用 dropLast()方法,但我一直收到错误

I'm creating a simple calculator app and currently struggling at deleting the last character when a my button is tapped. I'm using the dropLast() method but I keep getting the error


调用参数#1缺少参数

Missing Argument for parameter #1 in call



@IBAction func onDelPressed (button: UIButton!)  {
     runningNumber = runningNumber.characters.dropLast()
     currentLbl.text = runningNumber
}


推荐答案

Swift 4(附录)



在Swift中,您可以申请 dropLast ()直接在 String 实例上,不再调用 .characters 来访问 字符串的字符串

Swift 4 (Addendum)

In Swift, you can apply dropLast() directectly on the String instance, no longer invoking .characters to access a CharacterView of the String:

var runningNumber = "12345"
runningNumber = String(runningNumber.dropLast())
print(runningNumber) // 1234



Swift 3(原始答案)



我假设 runningNumber 是一个 String 实例。在这种情况下, runningNumber.characters.dropLast()的类型不是 String ,而是 CharacterView

Swift 3 (Original answer)

I'll assume runningNumber is a String instance. In this case, runningNumber.characters.dropLast() is not of type String, but a CharacterView:

let foo = runningNumber.characters.dropLast()
print(type(of: foo)) // CharacterView

你需要使用 CharacterView 实例化 String 实例,然后将其分配回类型 String 的属性,例如

You need to use the CharacterView to instantiate a String instance prior to assigning it back to a property of type String, e.g.

var runningNumber = "12345"
runningNumber = String(runningNumber.characters.dropLast())
print(runningNumber) // 1234

即你的情况

@IBAction func onDelPressed (button: UIButton!)  {
  runningNumber = String(runningNumber.characters.dropLast())
  currentLbl.text = runningNumber
}

这篇关于删除Swift 3中的最后一个字符的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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