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I have the following simple code written in Swift 3:
let str = "Hello, playground"
let index = str.index(of: ",")!
let newStr = str.substring(to: index)
From Xcode 9 beta 5, I get the following warning:
‘
substring(to:)‘ is deprecated: Please useStringslicing subscript with a ‘partial range from’ operator.
How can this slicing subscript with partial range from be used in Swift 4?
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asked Aug 8, 2017 at 8:04
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1
You should leave one side empty, hence the name “partial range”.
let newStr = str[..<index]
The same stands for partial range from operators, just leave the other side empty:
let newStr = str[index...]
Keep in mind that these range operators return a Substring. If you want to convert it to a string, use String‘s initialization function:
let newStr = String(str[..<index])
You can read more about the new substrings here.
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answered Aug 8, 2017 at 8:07
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9
Convert Substring (Swift 3) to String Slicing (Swift 4)
Examples In Swift 3, 4:
let newStr = str.substring(to: index) // Swift 3
let newStr = String(str[..<index]) // Swift 4
let newStr = str.substring(from: index) // Swift 3
let newStr = String(str[index...]) // Swift 4
let range = firstIndex..<secondIndex // If you have a range
let newStr = = str.substring(with: range) // Swift 3
let newStr = String(str[range]) // Swift 4
answered Oct 7, 2017 at 5:51
1
Swift 5, 4
Usage
let text = "Hello world"
text[0] // H
text[...3] // "Hell"
text[6..<text.count] // world
text[NSRange(location: 6, length: 3)] // wor
Code
import Foundation
public extension String {
subscript(value: Int) -> Character {
self[index(at: value)]
}
}
public extension String {
subscript(value: NSRange) -> Substring {
self[value.lowerBound..<value.upperBound]
}
}
public extension String {
subscript(value: CountableClosedRange<Int>) -> Substring {
self[index(at: value.lowerBound)...index(at: value.upperBound)]
}
subscript(value: CountableRange<Int>) -> Substring {
self[index(at: value.lowerBound)..<index(at: value.upperBound)]
}
subscript(value: PartialRangeUpTo<Int>) -> Substring {
self[..<index(at: value.upperBound)]
}
subscript(value: PartialRangeThrough<Int>) -> Substring {
self[...index(at: value.upperBound)]
}
subscript(value: PartialRangeFrom<Int>) -> Substring {
self[index(at: value.lowerBound)...]
}
}
private extension String {
func index(at offset: Int) -> String.Index {
index(startIndex, offsetBy: offset)
}
}
answered Oct 8, 2017 at 18:20
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Shorter in Swift 4/5:
let string = "123456"
let firstThree = String(string.prefix(3)) //"123"
let lastThree = String(string.suffix(3)) //"456"
answered Mar 23, 2018 at 10:44
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4
The conversion of your code to Swift 4 can also be done this way:
let str = "Hello, playground"
let index = str.index(of: ",")!
let substr = str.prefix(upTo: index)
You can use the code below to have a new string:
let newString = String(str.prefix(upTo: index))
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answered Sep 20, 2017 at 12:18
Swift5
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])
}
}
Usage:
var str = "Hello, Nick Michaels"
print(str.subString(from:7,to:20))
// print Nick Michaels
answered Aug 24, 2018 at 20:52
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substring(from: index) Converted to [index…]
Check the sample
let text = "1234567890"
let index = text.index(text.startIndex, offsetBy: 3)
text.substring(from: index) // "4567890" [Swift 3]
String(text[index...]) // "4567890" [Swift 4]
answered Sep 28, 2017 at 1:39
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Some useful extensions:
extension String {
func substring(from: Int, to: Int) -> String {
let start = index(startIndex, offsetBy: from)
let end = index(start, offsetBy: to - from)
return String(self[start ..< end])
}
func substring(range: NSRange) -> String {
return substring(from: range.lowerBound, to: range.upperBound)
}
}
answered Oct 3, 2017 at 8:20
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Example of uppercasedFirstCharacter convenience property in Swift3 and Swift4.
Property uppercasedFirstCharacterNew demonstrates how to use String slicing subscript in Swift4.
extension String {
public var uppercasedFirstCharacterOld: String {
if characters.count > 0 {
let splitIndex = index(after: startIndex)
let firstCharacter = substring(to: splitIndex).uppercased()
let sentence = substring(from: splitIndex)
return firstCharacter + sentence
} else {
return self
}
}
public var uppercasedFirstCharacterNew: String {
if characters.count > 0 {
let splitIndex = index(after: startIndex)
let firstCharacter = self[..<splitIndex].uppercased()
let sentence = self[splitIndex...]
return firstCharacter + sentence
} else {
return self
}
}
}
let lorem = "lorem".uppercasedFirstCharacterOld
print(lorem) // Prints "Lorem"
let ipsum = "ipsum".uppercasedFirstCharacterNew
print(ipsum) // Prints "Ipsum"
answered Aug 8, 2017 at 21:33
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You can create your custom subString method using extension to class String as below:
extension String {
func subString(startIndex: Int, endIndex: Int) -> String {
let end = (endIndex - self.count) + 1
let indexStartOfText = self.index(self.startIndex, offsetBy: startIndex)
let indexEndOfText = self.index(self.endIndex, offsetBy: end)
let substring = self[indexStartOfText..<indexEndOfText]
return String(substring)
}
}
answered Dec 26, 2017 at 15:48
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Creating SubString (prefix and suffix) from String using Swift 4:
let str : String = "ilike"
for i in 0...str.count {
let index = str.index(str.startIndex, offsetBy: i) // String.Index
let prefix = str[..<index] // String.SubSequence
let suffix = str[index...] // String.SubSequence
print("prefix \(prefix), suffix : \(suffix)")
}
Output
prefix , suffix : ilike
prefix i, suffix : like
prefix il, suffix : ike
prefix ili, suffix : ke
prefix ilik, suffix : e
prefix ilike, suffix :
If you want to generate a substring between 2 indices , use :
let substring1 = string[startIndex...endIndex] // including endIndex
let subString2 = string[startIndex..<endIndex] // excluding endIndex
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answered Dec 8, 2017 at 2:45
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I have written a string extension for replacement of ‘String: subString:’
extension String {
func sliceByCharacter(from: Character, to: Character) -> String? {
let fromIndex = self.index(self.index(of: from)!, offsetBy: 1)
let toIndex = self.index(self.index(of: to)!, offsetBy: -1)
return String(self[fromIndex...toIndex])
}
func sliceByString(from:String, to:String) -> String? {
//From - startIndex
var range = self.range(of: from)
let subString = String(self[range!.upperBound...])
//To - endIndex
range = subString.range(of: to)
return String(subString[..<range!.lowerBound])
}
}
Usage :
"Date(1511508780012+0530)".sliceByString(from: "(", to: "+")Example Result : “1511508780012”
PS: Optionals are forced to unwrap. Please add Type safety check wherever necessary.
answered Nov 24, 2017 at 12:39
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If you are trying to just get a substring up to a specific character, you don’t need to find the index first, you can just use the prefix(while:) method
let str = "Hello, playground"
let subString = str.prefix { $0 != "," } // "Hello" as a String.SubSequence
answered Jul 22, 2021 at 11:00
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When programming I often have strings with just plain A-Za-z and 0-9. No need for difficult Index actions. This extension is based on the plain old left / mid / right functions.
extension String {
// LEFT
// Returns the specified number of chars from the left of the string
// let str = "Hello"
// print(str.left(3)) // Hel
func left(_ to: Int) -> String {
return "\(self[..<self.index(startIndex, offsetBy: to)])"
}
// RIGHT
// Returns the specified number of chars from the right of the string
// let str = "Hello"
// print(str.left(3)) // llo
func right(_ from: Int) -> String {
return "\(self[self.index(startIndex, offsetBy: self.length-from)...])"
}
// MID
// Returns the specified number of chars from the startpoint of the string
// let str = "Hello"
// print(str.left(2,amount: 2)) // ll
func mid(_ from: Int, amount: Int) -> String {
let x = "\(self[self.index(startIndex, offsetBy: from)...])"
return x.left(amount)
}
}
answered Oct 13, 2017 at 17:28
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Hope this will help little more :-
var string = "123456789"
If you want a substring after some particular index.
var indexStart = string.index(after: string.startIndex )// you can use any index in place of startIndex
var strIndexStart = String (string[indexStart...])//23456789
If you want a substring after removing some string at the end.
var indexEnd = string.index(before: string.endIndex)
var strIndexEnd = String (string[..<indexEnd])//12345678
you can also create indexes with the following code :-
var indexWithOffset = string.index(string.startIndex, offsetBy: 4)
answered Feb 15, 2018 at 7:48
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with this method you can get specific range of string.you need to pass start index and after that total number of characters you want.
extension String{
func substring(fromIndex : Int,count : Int) -> String{
let startIndex = self.index(self.startIndex, offsetBy: fromIndex)
let endIndex = self.index(self.startIndex, offsetBy: fromIndex + count)
let range = startIndex..<endIndex
return String(self[range])
}
}
answered May 24, 2018 at 10:26
This is my solution, no warning, no errors, but perfect
let redStr: String = String(trimmStr[String.Index.init(encodedOffset: 0)..<String.Index.init(encodedOffset: 2)])
let greenStr: String = String(trimmStr[String.Index.init(encodedOffset: 3)..<String.Index.init(encodedOffset: 4)])
let blueStr: String = String(trimmStr[String.Index.init(encodedOffset: 5)..<String.Index.init(encodedOffset: 6)])
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answered Nov 7, 2017 at 5:37
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var str = "Hello, playground"
let indexcut = str.firstIndex(of: ",")
print(String(str[..<indexcut!]))
print(String(str[indexcut!...]))
You can try in this way and will get proper results.
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answered Jul 24, 2019 at 12:09
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Hope it would be helpful.
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(","))
answered May 31, 2018 at 12:10
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the simples way that I use is :
String(Array(str)[2...4])
answered Sep 21, 2021 at 13:06
0
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