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
在项目中包含此扩展
extension String{
func trim() -> String
{
return self.trimmingCharacters(in: NSCharacterSet.whitespaces)
}
var length: Int {
return self.count
}
subscript (i: Int) -> String {
return self[i ..< i + 1]
}
func substring(fromIndex: Int) -> String {
return self[min(fromIndex, length) ..< length]
}
func substring(toIndex: Int) -> String {
return self[0 ..< max(0, toIndex)]
}
subscript (r: Range<Int>) -> String {
let range = Range(uncheckedBounds: (lower: max(0, min(length, r.lowerBound)),
upper: min(length, max(0, r.upperBound))))
let start = index(startIndex, offsetBy: range.lowerBound)
let end = index(start, offsetBy: range.upperBound - range.lowerBound)
return String(self[start ..< end])
}
func substring(fromIndex: Int, toIndex:Int)->String{
let startIndex = self.index(self.startIndex, offsetBy: fromIndex)
let endIndex = self.index(startIndex, offsetBy: toIndex-fromIndex)
return String(self[startIndex...endIndex])
}
然后像这样使用函数
let str = "Sample-String"
let substring = str.substring(fromIndex: 0, toIndex: 0) //returns S
let sampleSubstr = str.substring(fromIndex: 0, toIndex: 5) //returns Sample
其他回答
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 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))
}
}
}
我的解决方案是在一行中，假设cadena是字符串，4是你想要的第n个位置:
let character = cadena[advance(cadena.startIndex, 4)]
简单的…我想Swift在未来的版本中会包含更多关于子字符串的内容。
我刚想出了一个巧妙的变通办法
var firstChar = Array(string)[0]
在项目中包含此扩展
extension String{
func trim() -> String
{
return self.trimmingCharacters(in: NSCharacterSet.whitespaces)
}
var length: Int {
return self.count
}
subscript (i: Int) -> String {
return self[i ..< i + 1]
}
func substring(fromIndex: Int) -> String {
return self[min(fromIndex, length) ..< length]
}
func substring(toIndex: Int) -> String {
return self[0 ..< max(0, toIndex)]
}
subscript (r: Range<Int>) -> String {
let range = Range(uncheckedBounds: (lower: max(0, min(length, r.lowerBound)),
upper: min(length, max(0, r.upperBound))))
let start = index(startIndex, offsetBy: range.lowerBound)
let end = index(start, offsetBy: range.upperBound - range.lowerBound)
return String(self[start ..< end])
}
func substring(fromIndex: Int, toIndex:Int)->String{
let startIndex = self.index(self.startIndex, offsetBy: fromIndex)
let endIndex = self.index(startIndex, offsetBy: toIndex-fromIndex)
return String(self[startIndex...endIndex])
}
然后像这样使用函数
let str = "Sample-String"
let substring = str.substring(fromIndex: 0, toIndex: 0) //returns S
let sampleSubstr = str.substring(fromIndex: 0, toIndex: 5) //returns Sample
