这是一个字符串扩展表单:

extension String {

    func removeCharsFromEnd(count_:Int) -> String {
        let stringLength = count(self)

        let substringIndex = (stringLength < count_) ? 0 : stringLength - count_

        return self.substringToIndex(advance(self.startIndex, substringIndex))
    }
}

对于Swift 1.2之前的版本:

...
let stringLength = countElements(self)
...

用法:

var str_1 = "Maxim"
println("output: \(str_1.removeCharsFromEnd(1))") // "Maxi"
println("output: \(str_1.removeCharsFromEnd(3))") // "Ma"
println("output: \(str_1.removeCharsFromEnd(8))") // ""

参考:

扩展向现有的类、结构或枚举类型添加新功能。这包括扩展您无法访问原始源代码的类型的能力(称为回溯建模)。扩展类似于Objective-C中的类别。(与Objective-C类别不同，Swift扩展没有名称。)

看文档

这是一个字符串扩展表单:

extension String {

    func removeCharsFromEnd(count_:Int) -> String {
        let stringLength = count(self)

        let substringIndex = (stringLength < count_) ? 0 : stringLength - count_

        return self.substringToIndex(advance(self.startIndex, substringIndex))
    }
}

对于Swift 1.2之前的版本:

...
let stringLength = countElements(self)
...

用法:

var str_1 = "Maxim"
println("output: \(str_1.removeCharsFromEnd(1))") // "Maxi"
println("output: \(str_1.removeCharsFromEnd(3))") // "Ma"
println("output: \(str_1.removeCharsFromEnd(8))") // ""

参考:

扩展向现有的类、结构或枚举类型添加新功能。这包括扩展您无法访问原始源代码的类型的能力(称为回溯建模)。扩展类似于Objective-C中的类别。(与Objective-C类别不同，Swift扩展没有名称。)

看文档

Swift 4.0(也叫Swift 5.0)

var str = "Hello, World"                           // "Hello, World"
str.dropLast()                                     // "Hello, Worl" (non-modifying)
str                                                // "Hello, World"
String(str.dropLast())                             // "Hello, Worl"

str.remove(at: str.index(before: str.endIndex))    // "d"
str                                                // "Hello, Worl" (modifying)

斯威夫特3.0

api变得更加敏捷了，因此Foundation扩展也做了一些改变:

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

或者就地版本:

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

谢谢Zmey, Rob Allen!

Swift 2.0+方式

有几种方法可以做到这一点:

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(改变名称):

var name: String = "Dolphin"    
name.removeAtIndex(name.endIndex.predecessor())
print(name) // "Dolphi"

使用dropLast()函数:

var name: String = "Dolphin"
var truncated = String(name.characters.dropLast())
print(name)      // "Dolphin"
print(truncated) // "Dolphi"

旧的字符串。索引(Xcode 6 Beta 4 +)方式

由于Swift中的字符串类型旨在提供出色的UTF-8支持，您不能再使用Int类型访问字符索引/范围/子字符串。相反，您使用String。指数:

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"

或者(对于一个更实用，但教育意义更小的例子)，你可以使用endIndex:

let name: String = "Dolphin"
name.substringToIndex(name.endIndex.predecessor()) // "Dolphi"

注意:我发现这是理解String的一个很好的起点。指数

旧(pre-Beta 4)方式

你可以简单地使用substringToIndex()函数，提供它比String的长度小1:

let name: String = "Dolphin"
name.substringToIndex(countElements(name) - 1) // "Dolphi"

斯威夫特4/5

var str = "bla"
str.removeLast() // returns "a"; str is now "bl"

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
    }
}

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"