从字符串中删除最后一个字 迅捷的语言

Question

如何使用Swift从String变量中删除最后一个字符？在文档中找不到它.

这是完整的例子:

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

Answer 1

Swift 4.0

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)

Swift 3.0

API变得更加开放,因此Foundation扩展已经改变了一点:

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

有几种方法可以实现这一目标:

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"

使用该removeRange()方法(改变了name):

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

使用dropLast()功能:

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

Old String.Index(Xcode 6 Beta 4 +)方式

由于StringSwift中的类型旨在提供出色的UTF-8支持,因此您无法再使用Int类型访问字符索引/范围/子字符串.相反,你使用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"

或者(对于更实际,但教育程度较低的示例),您可以使用endIndex:

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

注意:我发现这是理解的一个很好的起点String.Index

旧(Beta 4之前)的方式

你可以简单地使用这个substringToIndex()函数,比它的长度少一个String:

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

Answer 2

全局dropLast()函数适用于序列,因此也适用于字符串:

var expression  = "45+22"
expression = dropLast(expression)  // "45+2"

// in Swift 2.0 (according to cromanelli's comment below)
expression = String(expression.characters.dropLast())

Answer 3

斯威夫特4:

let choppedString = String(theString.dropLast())

在Swift 2中,执行以下操作:

let choppedString = String(theString.characters.dropLast())

我建议使用此链接来了解Swift字符串.

Answer 4

斯威夫特 4/5

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

Answer 5

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))
    }
}

对于早于1.2的Swift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:

扩展为现有的类,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(称为追溯建模).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.)

见DOCS

Answer 6

使用功能removeAtIndex(i: String.Index) -> Character:

var s = "abc"    
s.removeAtIndex(s.endIndex.predecessor())  // "ab"

Answer 7

修剪字符串最后一个字符的最简单方法是:

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

Answer 8

斯威夫特4

var welcome = "Hello World!"
welcome = String(welcome[..<welcome.index(before:welcome.endIndex)])

要么

welcome.remove(at: welcome.index(before: welcome.endIndex))

要么

welcome = String(welcome.dropLast())