问题描述
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)
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!
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
):
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"
由于String
Swift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
:
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
:
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
您可以简单地使用该substringToIndex()
函数,为其提供的长度要小于String
:
let name: String = "Dolphin"
name.substringToIndex(countElements(name) - 1) // "Dolphi"
解决方法
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)