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
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 "" }
...
}
其他回答
我想指出的是,如果你有一个很大的字符串,并且需要从中随机访问许多字符,你可能想要支付额外的内存成本,并将字符串转换为一个数组以获得更好的性能:
// Pay up front for O(N) memory
let chars = Array(veryLargeString.characters)
for i in 0...veryLargeNumber {
// Benefit from O(1) access
print(chars[i])
}
斯威夫特3
extension String {
public func charAt(_ i: Int) -> Character {
return self[self.characters.index(self.startIndex, offsetBy: i)]
}
public subscript (i: Int) -> String {
return String(self.charAt(i) as Character)
}
public subscript (r: Range<Int>) -> String {
return substring(with: self.characters.index(self.startIndex, offsetBy: r.lowerBound)..<self.characters.index(self.startIndex, offsetBy: r.upperBound))
}
public subscript (r: CountableClosedRange<Int>) -> String {
return substring(with: self.characters.index(self.startIndex, offsetBy: r.lowerBound)..<self.characters.index(self.startIndex, offsetBy: r.upperBound))
}
}
使用
let str = "Hello World"
let sub = str[0...4]
有用的编程技巧和技巧(我写的)
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]
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]
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