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 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
其他回答
斯威夫特4
let str = "My String"
索引处的字符串
let index = str.index(str.startIndex, offsetBy: 3)
String(str[index]) // "S"
子字符串
let startIndex = str.index(str.startIndex, offsetBy: 3)
let endIndex = str.index(str.startIndex, offsetBy: 7)
String(str[startIndex...endIndex]) // "Strin"
前n个字符
let startIndex = str.index(str.startIndex, offsetBy: 3)
String(str[..<startIndex]) // "My "
最后n个字符
let startIndex = str.index(str.startIndex, offsetBy: 3)
String(str[startIndex...]) // "String"
Swift 2和3
str = "My String"
**字符串索引**
斯威夫特2
let charAtIndex = String(str[str.startIndex.advancedBy(3)]) // charAtIndex = "S"
斯威夫特3
str[str.index(str.startIndex, offsetBy: 3)]
子字符串fromIndex toIndex
斯威夫特2
let subStr = str[str.startIndex.advancedBy(3)...str.startIndex.advancedBy(7)] // subStr = "Strin"
斯威夫特3
str[str.index(str.startIndex, offsetBy: 3)...str.index(str.startIndex, offsetBy: 7)]
前n个字符
let first2Chars = String(str.characters.prefix(2)) // first2Chars = "My"
最后n个字符
let last3Chars = String(str.characters.suffix(3)) // last3Chars = "ing"
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”
在Swift 3中,没有对String类的扩展,就像我能做的那样简单!
let myString = "abcedfg"
let characterLocationIndex = myString.index(myString.startIndex, offsetBy: 3)
let myCharacter = myString[characterLocationIndex]
本例中的myCharacter为“3”。
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])
允许负指数
它总是有用的,不必总是写string[string]。长度- 1]用于在使用下标扩展名时获取最后一个字符。这(Swift 3)扩展允许负索引,范围和CountableClosedRange。
extension String {
var count: Int { return self.characters.count }
subscript (i: Int) -> Character {
// wraps out of bounds indices
let j = i % self.count
// wraps negative indices
let x = j < 0 ? j + self.count : j
// quick exit for first
guard x != 0 else {
return self.characters.first!
}
// quick exit for last
guard x != count - 1 else {
return self.characters.last!
}
return self[self.index(self.startIndex, offsetBy: x)]
}
subscript (r: Range<Int>) -> String {
let lb = r.lowerBound
let ub = r.upperBound
// quick exit for one character
guard lb != ub else { return String(self[lb]) }
return self[self.index(self.startIndex, offsetBy: lb)..<self.index(self.startIndex, offsetBy: ub)]
}
subscript (r: CountableClosedRange<Int>) -> String {
return self[r.lowerBound..<r.upperBound + 1]
}
}
如何使用:
var text = "Hello World"
text[-1] // d
text[2] // l
text[12] // e
text[0...4] // Hello
text[0..<4] // Hell
对于更彻底的程序员:在这个扩展中包括一个防止空字符串的保护
subscript (i: Int) -> Character {
guard self.count != 0 else { return '' }
...
}
subscript (r: Range<Int>) -> String {
guard self.count != 0 else { return "" }
...
}