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
一个类似蟒蛇的解决方案,允许你使用负下标,
var str = "Hello world!"
str[-1] // "!"
可能是:
extension String {
subscript (var index:Int)->Character{
get {
let n = distance(self.startIndex, self.endIndex)
index %= n
if index < 0 { index += n }
return self[advance(startIndex, index)]
}
}
}
顺便说一下,调换整个python的切片符号可能是值得的
其他回答
到目前为止,下标(_:)不可用。就像我们不能这么做一样
str[0]
与字符串。我们必须提供"String "。指数” 但是,我们如何给出我们自己的索引号呢,我们可以用,
string[str.index(str.startIndex, offsetBy: 0)]
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
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]
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))
}
}
}
在Swift 5中,不扩展字符串:
var str = "ABCDEFGH"
for char in str {
if(char == "C") { }
}
以上Swift代码与Java代码相同:
int n = 8;
var str = "ABCDEFGH"
for (int i=0; i<n; i++) {
if (str.charAt(i) == 'C') { }
}