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"
其他回答
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 ???
你可以通过将字符串转换为数组,并通过下标获得特定的索引,如下所示
var str = "Hello"
let s = Array(str)[2]
print(s)
对我来说最有效的方法是:
var firstName = "Olivia"
var lastName = "Pope"
var nameInitials.text = "\(firstName.prefix(1))" + "\ (lastName.prefix(1))"
输出:“OP”
Swift 4.2或更高版本
使用String的indexes属性进行范围和部分范围下标
作为@LeoDabus nice answer的变体,我们可以为DefaultIndices添加一个额外的扩展,目的是允许我们在为后者实现自定义下标(通过Int专用范围和部分范围)时使用String的indexes属性。
extension DefaultIndices {
subscript(at: Int) -> Elements.Index { index(startIndex, offsetBy: at) }
}
// Moving the index(_:offsetBy:) to an extension yields slightly
// briefer implementations for these String extensions.
extension String {
subscript(range: Range<Int>) -> SubSequence {
let start = indices[range.lowerBound]
return self[start..<indices[start...][range.count]]
}
subscript(range: ClosedRange<Int>) -> SubSequence {
let start = indices[range.lowerBound]
return self[start...indices[start...][range.count]]
}
subscript(range: PartialRangeFrom<Int>) -> SubSequence {
self[indices[range.lowerBound]...]
}
subscript(range: PartialRangeThrough<Int>) -> SubSequence {
self[...indices[range.upperBound]]
}
subscript(range: PartialRangeUpTo<Int>) -> SubSequence {
self[..<indices[range.upperBound]]
}
}
let str = "foo bar baz bax"
print(str[4..<6]) // "ba"
print(str[4...6]) // "bar"
print(str[4...]) // "bar baz bax"
print(str[...6]) // "foo bar"
print(str[..<6]) // "foo ba"
感谢@LeoDabus指出我在使用索引属性作为字符串下标的(其他)替代方案的方向!
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"
:-)