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5.1
这可能是所有答案中最简单的一个。
添加这个扩展:
extension String {
func retrieveFirstCharacter() -> String? {
guard self.count > 0 else { return nil }
let numberOfCharacters = self.count
return String(self.dropLast(numberOfCharacters - 1))
}
}
其他回答
swift 2.0子字符串更新
public extension String {
public subscript (i: Int) -> String {
return self.substringWithRange(self.startIndex..<self.startIndex.advancedBy(i + 1))
}
public subscript (r: Range<Int>) -> String {
get {
return self.substringWithRange(self.startIndex.advancedBy(r.startIndex)..<self.startIndex.advancedBy(r.endIndex))
}
}
}
我的解决方法很简单:
斯威夫特4.1:
let myString = "Test string"
let index = 0
let firstCharacter = myString[String.Index(encodedOffset: index)]
斯威夫特5.1:
let firstCharacter = myString[String.Index.init(utf16Offset: index, in: myString)]
不使用整数进行索引,只使用String.Index。主要是线性复杂度。您还可以从String中创建范围。索引并使用它们获取子字符串。
斯威夫特3.0
let firstChar = someString[someString.startIndex]
let lastChar = someString[someString.index(before: someString.endIndex)]
let charAtIndex = someString[someString.index(someString.startIndex, offsetBy: 10)]
let range = someString.startIndex..<someString.index(someString.startIndex, offsetBy: 10)
let substring = someString[range]
快2.倍
let firstChar = someString[someString.startIndex]
let lastChar = someString[someString.endIndex.predecessor()]
let charAtIndex = someString[someString.startIndex.advanceBy(10)]
let range = someString.startIndex..<someString.startIndex.advanceBy(10)
let subtring = someString[range]
请注意,不能使用从一个字符串到另一个字符串创建的索引(或范围)
let index10 = someString.startIndex.advanceBy(10)
//will compile
//sometimes it will work but sometimes it will crash or result in undefined behaviour
let charFromAnotherString = anotherString[index10]
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)!
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"