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 2.0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))
}
}
}
其他回答
Swift的String类型没有提供characterAtIndex方法,因为Unicode字符串有几种编码方式。你要用UTF8, UTF16,还是别的?
您可以通过检索String来访问CodeUnit集合。utf8和String。utf16属性。您还可以通过检索String来访问UnicodeScalar集合。unicodeScalars财产。
在NSString实现的精神中,我返回一个unichar类型。
extension String
{
func characterAtIndex(index:Int) -> unichar
{
return self.utf16[index]
}
// Allows us to use String[index] notation
subscript(index:Int) -> unichar
{
return characterAtIndex(index)
}
}
let text = "Hello Swift!"
let firstChar = text[0]
我是斯威夫特4号
let myString = "LOVE"
self.textField1.text = String(Array(myString)[0])
self.textField2.text = String(Array(myString)[1])
self.textField3.text = String(Array(myString)[2])
self.textField4.text = String(Array(myString)[3])
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。
为了提供主题并显示快速下标的可能性,这里有一个基于“substring-toolbox”下标的小字符串
这些方法是安全的,并且不会遍历字符串索引
extension String {
// string[i] -> one string char
subscript(pos: Int) -> String { return String(Array(self)[min(self.length-1,max(0,pos))]) }
// string[pos,len] -> substring from pos for len chars on the left
subscript(pos: Int, len: Int) -> String { return self[pos, len, .pos_len, .left2right] }
// string[pos, len, .right2left] -> substring from pos for len chars on the right
subscript(pos: Int, len: Int, way: Way) -> String { return self[pos, len, .pos_len, way] }
// string[range] -> substring form start pos on the left to end pos on the right
subscript(range: Range<Int>) -> String { return self[range.startIndex, range.endIndex, .start_end, .left2right] }
// string[range, .right2left] -> substring start pos on the right to end pos on the left
subscript(range: Range<Int>, way: Way) -> String { return self[range.startIndex, range.endIndex, .start_end, way] }
var length: Int { return countElements(self) }
enum Mode { case pos_len, start_end }
enum Way { case left2right, right2left }
subscript(var val1: Int, var val2: Int, mode: Mode, way: Way) -> String {
if mode == .start_end {
if val1 > val2 { let val=val1 ; val1=val2 ; val2=val }
val2 = val2-val1
}
if way == .left2right {
val1 = min(self.length-1, max(0,val1))
val2 = min(self.length-val1, max(1,val2))
} else {
let val1_ = val1
val1 = min(self.length-1, max(0, self.length-val1_-val2 ))
val2 = max(1, (self.length-1-val1_)-(val1-1) )
}
return self.bridgeToObjectiveC().substringWithRange(NSMakeRange(val1, val2))
//-- Alternative code without bridge --
//var range: Range<Int> = pos...(pos+len-1)
//var start = advance(startIndex, range.startIndex)
//var end = advance(startIndex, range.endIndex)
//return self.substringWithRange(Range(start: start, end: end))
}
}
println("0123456789"[3]) // return "3"
println("0123456789"[3,2]) // return "34"
println("0123456789"[3,2,.right2left]) // return "56"
println("0123456789"[5,10,.pos_len,.left2right]) // return "56789"
println("0123456789"[8,120,.pos_len,.right2left]) // return "01"
println("0123456789"[120,120,.pos_len,.left2right]) // return "9"
println("0123456789"[0...4]) // return "01234"
println("0123456789"[0..4]) // return "0123"
println("0123456789"[0...4,.right2left]) // return "56789"
println("0123456789"[4...0,.right2left]) // return "678" << because ??? range can wear endIndex at 0 ???
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)]