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。
这是完整的示例:
var expression = "45+22" expression = expression.substringToIndex(countElements(expression) - 1)
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)
斯威夫特3.0
这些API变得更加 快捷 ,因此Foundation扩展有所改变:
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"
感谢Zmey,Rob Allen!
Swift 2.0+方式
有几种方法可以实现此目的:
通过Foundation扩展,尽管不属于Swift库:
var name: String = "Dolphin" var truncated = name.substringToIndex(name.endIndex.predecessor()) print(name) // "Dolphin" print(truncated) // "Dolphi"
使用removeRange()方法(其 涂改 的name):
removeRange()
name
var name: String = "Dolphin" name.removeAtIndex(name.endIndex.predecessor()) print(name) // "Dolphi"
使用dropLast()功能:
dropLast()
var name: String = "Dolphin" var truncated = String(name.characters.dropLast()) print(name) // "Dolphin" print(truncated) // "Dolphi"
旧的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,因此您不能再使用Int类型访问字符索引/范围/子字符串。而是使用String.Index:
String
Int
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"
另外(对于一个更实际,但不太教育的示例),您可以使用endIndex:
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
旧版(测试版4之前)
您可以简单地使用该substringToIndex()函数,为其提供的长度要小于String:
substringToIndex()
let name: String = "Dolphin" name.substringToIndex(countElements(name) - 1) // "Dolphi"