我怎样才能得到字符串的第n个字符?我尝试了括号([])访问器,没有运气。
var string = "Hello, world!"
var firstChar = string[0] // Throws error
错误:'下标'是不可用的:不能下标String与Int,请参阅文档注释讨论
我怎样才能得到字符串的第n个字符?我尝试了括号([])访问器,没有运气。
var string = "Hello, world!"
var firstChar = string[0] // Throws error
错误:'下标'是不可用的:不能下标String与Int,请参阅文档注释讨论
当前回答
Swift 2.2解决方案:
下面的扩展在Xcode 7中工作,这是这个解决方案和Swift 2.0语法转换的组合。
extension String {
subscript(integerIndex: Int) -> Character {
let index = startIndex.advancedBy(integerIndex)
return self[index]
}
subscript(integerRange: Range<Int>) -> String {
let start = startIndex.advancedBy(integerRange.startIndex)
let end = startIndex.advancedBy(integerRange.endIndex)
let range = start..<end
return self[range]
}
}
其他回答
顺便说一句,有几个函数可以直接应用于String的字符链表示,像这样:
var string = "Hello, playground"
let firstCharacter = string.characters.first // returns "H"
let lastCharacter = string.characters.last // returns "d"
结果类型为Character,但可以将其转换为String。
或:
let reversedString = String(string.characters.reverse())
// returns "dnuorgyalp ,olleH"
:-)
如果你看到不能下标一个类型为'String'的值…使用这个扩展:
斯威夫特3
extension String {
subscript (i: Int) -> Character {
return self[self.characters.index(self.startIndex, offsetBy: i)]
}
subscript (i: Int) -> String {
return String(self[i] as Character)
}
subscript (r: Range<Int>) -> String {
let start = index(startIndex, offsetBy: r.lowerBound)
let end = index(startIndex, offsetBy: r.upperBound)
return self[start..<end]
}
subscript (r: ClosedRange<Int>) -> String {
let start = index(startIndex, offsetBy: r.lowerBound)
let end = index(startIndex, offsetBy: r.upperBound)
return self[start...end]
}
}
斯威夫特2.3
extension String {
subscript(integerIndex: Int) -> Character {
let index = advance(startIndex, integerIndex)
return self[index]
}
subscript(integerRange: Range<Int>) -> String {
let start = advance(startIndex, integerRange.startIndex)
let end = advance(startIndex, integerRange.endIndex)
let range = start..<end
return self[range]
}
}
来源:http://oleb.net/blog/2014/07/swift-strings/
斯威夫特4.2
这个答案是理想的,因为它在一个扩展中扩展了String及其所有子序列(Substring)
public extension StringProtocol {
public subscript (i: Int) -> Element {
return self[index(startIndex, offsetBy: i)]
}
public subscript (bounds: CountableClosedRange<Int>) -> SubSequence {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
return self[start...end]
}
public subscript (bounds: CountableRange<Int>) -> SubSequence {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
return self[start..<end]
}
public subscript (bounds: PartialRangeUpTo<Int>) -> SubSequence {
let end = index(startIndex, offsetBy: bounds.upperBound)
return self[startIndex..<end]
}
public subscript (bounds: PartialRangeThrough<Int>) -> SubSequence {
let end = index(startIndex, offsetBy: bounds.upperBound)
return self[startIndex...end]
}
public subscript (bounds: CountablePartialRangeFrom<Int>) -> SubSequence {
let start = index(startIndex, offsetBy: bounds.lowerBound)
return self[start..<endIndex]
}
}
使用
var str = "Hello, playground"
print(str[5...][...5][0])
// Prints ","
使用字符就可以了。您可以快速地将字符串转换为字符数组,可以由CharacterView方法操作。
例子:
let myString = "Hello World!"
let myChars = myString.characters
(完整的CharacterView文档)
(在Swift 3中测试)
我刚想出了一个巧妙的变通办法
var firstChar = Array(string)[0]