我有以下简单的代码写在Swift 3:
let str = "Hello, playground"
let index = str.index(of: ",")!
let newStr = str.substring(to: index)
在Xcode 9 beta 5中,我得到了以下警告:
'substring(to:)'已弃用:请使用带有'partial range from'操作符的字符串切片下标。
这个部分范围的切片下标如何在Swift 4中使用?
我有以下简单的代码写在Swift 3:
let str = "Hello, playground"
let index = str.index(of: ",")!
let newStr = str.substring(to: index)
在Xcode 9 beta 5中,我得到了以下警告:
'substring(to:)'已弃用:请使用带有'partial range from'操作符的字符串切片下标。
这个部分范围的切片下标如何在Swift 4中使用?
当前回答
Swift 4/5更短:
let string = "123456"
let firstThree = String(string.prefix(3)) //"123"
let lastThree = String(string.suffix(3)) //"456"
其他回答
希望这将帮助更多:-
var string = "123456789"
如果你想在某个特定的索引后面加一个子字符串。
var indexStart = string.index(after: string.startIndex )// you can use any index in place of startIndex
var strIndexStart = String (string[indexStart...])//23456789
如果你想在末尾删除某个字符串后获得子字符串。
var indexEnd = string.index(before: string.endIndex)
var strIndexEnd = String (string[..<indexEnd])//12345678
还可以使用以下代码创建索引:—
var indexWithOffset = string.index(string.startIndex, offsetBy: 4)
希望对大家有所帮助。
extension String {
func getSubString(_ char: Character) -> String {
var subString = ""
for eachChar in self {
if eachChar == char {
return subString
} else {
subString += String(eachChar)
}
}
return subString
}
}
let str: String = "Hello, playground"
print(str.getSubString(","))
Swift 4/5更短:
let string = "123456"
let firstThree = String(string.prefix(3)) //"123"
let lastThree = String(string.suffix(3)) //"456"
Swift5
(Java的子字符串方法):
extension String {
func subString(from: Int, to: Int) -> String {
let startIndex = self.index(self.startIndex, offsetBy: from)
let endIndex = self.index(self.startIndex, offsetBy: to)
return String(self[startIndex..<endIndex])
}
}
用法:
var str = "Hello, Nick Michaels"
print(str.subString(from:7,to:20))
// print Nick Michaels
你的代码转换到Swift 4也可以这样做:
let str = "Hello, playground"
let index = str.index(of: ",")!
let substr = str.prefix(upTo: index)
你可以使用下面的代码来创建一个新的字符串:
let newString = String(str.prefix(upTo: index))