我如何从字符串变量使用Swift删除最后一个字符?在文档中找不到。

下面是完整的例子:

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

当前回答

斯威夫特4.2

我还删除了我的最后一个字符从字符串(即UILabel文本)在IOS应用程序

@IBOutlet weak var labelText: UILabel! // Do Connection with UILabel

@IBAction func whenXButtonPress(_ sender: UIButton) { // Do Connection With X Button

    labelText.text = String((labelText.text?.dropLast())!) // Delete the last caracter and assign it

}

其他回答

我更喜欢下面的实现,因为我不必担心,即使字符串是空的

let str = "abc"
str.popLast()

// Prints ab

str = ""
str.popLast() // It returns the Character? which is an optional

// Print <emptystring>

修剪字符串最后一个字符最简单的方法是:

title = title[title.startIndex ..< title.endIndex.advancedBy(-1)]

一个快速变化的类别:

extension String {
    mutating func removeCharsFromEnd(removeCount:Int)
    {
        let stringLength = count(self)
        let substringIndex = max(0, stringLength - removeCount)
        self = self.substringToIndex(advance(self.startIndex, substringIndex))
    }
}

使用:

var myString = "abcd"
myString.removeCharsFromEnd(2)
println(myString) // "ab"

简单回答(2015-04-16有效):removeAtIndex(mystring . endindex .前任())

例子:

var howToBeHappy = "Practice compassion, attention and gratitude. And smile!!"
howToBeHappy.removeAtIndex(howToBeHappy.endIndex.predecessor())
println(howToBeHappy)
// "Practice compassion, attention and gratitude. And smile!"

元:

语言继续着它的快速进化,使得许多以前很好的sos答案的半衰期变得危险地短暂。学习语言并参考真正的文档总是最好的。

Swift 3:当你想删除尾随字符串:

func replaceSuffix(_ suffix: String, replacement: String) -> String {
    if hasSuffix(suffix) {
        let sufsize = suffix.count < count ? -suffix.count : 0
        let toIndex = index(endIndex, offsetBy: sufsize)
        return substring(to: toIndex) + replacement
    }
    else
    {
        return self
    }
}