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
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 ","
其他回答
Swift3
可以使用下标语法访问特定String索引处的Character。
let greeting = "Guten Tag!"
let index = greeting.index(greeting.startIndex, offsetBy: 7)
greeting[index] // a
访问https://developer.apple.com/library/content/documentation/Swift/Conceptual/Swift_Programming_Language/StringsAndCharacters.html
或者我们可以在Swift 4中做一个字符串扩展
extension String {
func getCharAtIndex(_ index: Int) -> Character {
return self[self.index(self.startIndex, offsetBy: index)]
}
}
用法:
let foo = "ABC123"
foo.getCharAtIndex(2) //C
斯威夫特3:
extension String {
func substring(fromPosition: UInt, toPosition: UInt) -> String? {
guard fromPosition <= toPosition else {
return nil
}
guard toPosition < UInt(characters.count) else {
return nil
}
let start = index(startIndex, offsetBy: String.IndexDistance(fromPosition))
let end = index(startIndex, offsetBy: String.IndexDistance(toPosition) + 1)
let range = start..<end
return substring(with: range)
}
}
"ffaabbcc".substring(fromPosition: 2, toPosition: 5) // return "aabb"
swift字符串类不提供在特定索引处获取字符的能力,因为它本身支持UTF字符。UTF字符在内存中的可变长度使得直接跳转到字符是不可能的。这意味着每次都必须手动遍历字符串。
您可以扩展String以提供一个方法,该方法将遍历字符,直到您需要的索引
extension String {
func characterAtIndex(index: Int) -> Character? {
var cur = 0
for char in self {
if cur == index {
return char
}
cur++
}
return nil
}
}
myString.characterAtIndex(0)!
Swift 3:另一个解决方案(在操场测试)
extension String {
func substr(_ start:Int, length:Int=0) -> String? {
guard start > -1 else {
return nil
}
let count = self.characters.count - 1
guard start <= count else {
return nil
}
let startOffset = max(0, start)
let endOffset = length > 0 ? min(count, startOffset + length - 1) : count
return self[self.index(self.startIndex, offsetBy: startOffset)...self.index(self.startIndex, offsetBy: endOffset)]
}
}
用法:
let txt = "12345"
txt.substr(-1) //nil
txt.substr(0) //"12345"
txt.substr(0, length: 0) //"12345"
txt.substr(1) //"2345"
txt.substr(2) //"345"
txt.substr(3) //"45"
txt.substr(4) //"5"
txt.substr(6) //nil
txt.substr(0, length: 1) //"1"
txt.substr(1, length: 1) //"2"
txt.substr(2, length: 1) //"3"
txt.substr(3, length: 1) //"4"
txt.substr(3, length: 2) //"45"
txt.substr(3, length: 3) //"45"
txt.substr(4, length: 1) //"5"
txt.substr(4, length: 2) //"5"
txt.substr(5, length: 1) //nil
txt.substr(5, length: -1) //nil
txt.substr(-1, length: -1) //nil
这是一个你可以使用的扩展,与Swift 3.1一起工作。单个索引将返回一个字符,这在索引字符串时似乎很直观,而范围将返回一个字符串。
extension String {
subscript (i: Int) -> Character {
return Array(self.characters)[i]
}
subscript (r: CountableClosedRange<Int>) -> String {
return String(Array(self.characters)[r])
}
subscript (r: CountableRange<Int>) -> String {
return self[r.lowerBound...r.upperBound-1]
}
}
扩展的一些例子:
let string = "Hello"
let c1 = string[1] // Character "e"
let c2 = string[-1] // fatal error: Index out of range
let r1 = string[1..<4] // String "ell"
let r2 = string[1...4] // String "ello"
let r3 = string[1...5] // fatal error: Array index is out of range
如果需要的话,你可以在上面的扩展中添加一个额外的方法来返回一个包含单个字符的String:
subscript (i: Int) -> String {
return String(self[i])
}
注意,在索引字符串时,你必须显式地指定你想要的类型:
let c: Character = string[3] // Character "l"
let s: String = string[0] // String "H"