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。
以下是完整示例:
var expression = "45+22"
expression = expression.substringToIndex(countElements(expression) - 1)
答案 0 :(得分:485)
Swift 4.0(也是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)
Swift 3.0
API已经有了更多 swifty ,因此基础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+ Way
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removeRange()
方法(更改 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"
旧String.Index(Xcode 6 Beta 4 +)方式
由于Swift中的String
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,您使用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this是理解String.Index
旧(Beta 4之前)方式
你可以简单地使用substringToIndex()
函数,比String
的长度少一个:
let name: String = "Dolphin"
name.substringToIndex(countElements(name) - 1) // "Dolphi"
答案 1 :(得分:88)
全局dropLast()
函数适用于序列,因此适用于字符串:
var expression = "45+22"
expression = dropLast(expression) // "45+2"
// in Swift 2.0 (according to cromanelli's comment below)
expression = String(expression.characters.dropLast())
答案 2 :(得分:67)
let choppedString = String(theString.dropLast())
在Swift 2中,执行此操作:
let choppedString = String(theString.characters.dropLast())
我建议使用this链接来了解Swift字符串。
答案 3 :(得分:8)
这是一个String 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:的
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:
扩展程序为现有的类,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。)
请参阅DOCS
答案 4 :(得分:6)
使用函数removeAtIndex(i: String.Index) -> Character
:
var s = "abc"
s.removeAtIndex(s.endIndex.predecessor()) // "ab"
答案 5 :(得分:5)
Swift 4
var welcome = "Hello World!"
welcome = String(welcome[..<welcome.index(before:welcome.endIndex)])
或
welcome.remove(at: welcome.index(before: welcome.endIndex))
或
welcome = String(welcome.dropLast())
答案 6 :(得分:5)
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)]
答案 7 :(得分:2)
let str = "abc"
let substr = str.substringToIndex(str.endIndex.predecessor()) // "ab"
答案 8 :(得分:2)
使用新的Substring类型:
var before: String = "Hello world!"
var lastCharIndex: Int = before.endIndex
var after:String = String(before[..<lastCharIndex])
print(after) // Hello world
缩短方式:
var before: String = "Hello world!"
after = String(before[..<before.endIndex])
print(after) // Hello world
答案 9 :(得分:2)
简答程(自2015-04-16起生效): removeAtIndex(myString.endIndex.predecessor())
示例:
var howToBeHappy = "Practice compassion, attention and gratitude. And smile!!"
howToBeHappy.removeAtIndex(howToBeHappy.endIndex.predecessor())
println(howToBeHappy)
// "Practice compassion, attention and gratitude. And smile!"
元:
这种语言继续快速发展,使许多以前很好的S.O.的半衰期延长。答案非常简短。最好是学习语言并参考真实的documentation。
答案 10 :(得分:2)
var str = "Hello, playground"
extension String {
var stringByDeletingLastCharacter: String {
return dropLast(self)
}
}
println(str.stringByDeletingLastCharacter) // "Hello, playgroun"
答案 11 :(得分:1)
一个变异的快速类别:
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"
答案 12 :(得分:1)
另一种方式如果您想从结尾删除一个或多个而不是一个字符。
var myStr = "Hello World!"
myStr = (myStr as NSString).substringToIndex((myStr as NSString).length-XX)
XX 是您要删除的字符数。
答案 13 :(得分:1)
Swift 3(根据docs)2016年11月20日
let range = expression.index(expression.endIndex, offsetBy: -numberOfCharactersToRemove)..<expression.endIndex
expression.removeSubrange(range)
答案 14 :(得分:1)
使用函数advance(startIndex, endIndex)
:
var str = "45+22"
str = str.substringToIndex(advance(str.startIndex, countElements(str) - 1))
答案 15 :(得分:0)
$ docker run --network=host jupyter/pyspark-notebook
答案 16 :(得分:0)
import UIKit
var str1 = "Hello, playground"
str1.removeLast()
print(str1)
var str2 = "Hello, playground"
str2.removeLast(3)
print(str2)
var str3 = "Hello, playground"
str3.removeFirst(2)
print(str3)
Output:-
Hello, playgroun
Hello, playgro
llo, playground
答案 17 :(得分:0)
Swift 4.2
我还从IOS应用程序中的字符串(即 UILabel文本)中删除了我的最后一个字符
@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
}
答案 18 :(得分:0)
快捷键3 :要删除结尾的字符串:
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
}
}
答案 19 :(得分:0)
dropLast()
函数删除字符串的最后一个元素。
var expression = "45+22"
expression = expression.dropLast()
答案 20 :(得分:0)
与上面的代码互补我想删除字符串的开头,无法在任何地方找到引用。我是这样做的:
var mac = peripheral.identifier.description
let range = mac.startIndex..<mac.endIndex.advancedBy(-50)
mac.removeRange(range) // trim 17 characters from the beginning
let txPower = peripheral.advertisements.txPower?.description
这会从字符串的开头剪掉17个字符(总字符串长度为67,我们从结尾开始-50,你就拥有它。
答案 21 :(得分:0)
我建议将NSString用于您想要操作的字符串。实际上把它想象成一个开发人员我从未遇到过Swift String会解决的NSString问题......我理解这些细微之处。但我还没有真正需要它们。
var foo = "Foo" as NSString
或
var foo: NSString = "blah"
或
// check bounds before you do this, e.g. foo.length > 0
// Note shortFoo is of type NSString
var shortFoo = foo.substringToIndex(foo.length-1)
然后整个世界的简单NSString字符串操作对您开放。
作为问题的答案
-(void)funcName: (const NSString *)name parameter: (void(^)(ClassName *input)) obj