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 useString
slicing subscript with a 'partial range from' operator.
How can this slicing subscript with partial range from be used in Swift 4?
In Swift 4 you slice a string into a substring using subscripting. The use of substring(from:) , substring(to:) and substring(with:) are all deprecated.
In Swift, the first property is used to return the first character of a string.
Swift – String Length/Count To get the length of a String in Swift, use count property of the string. count property is an integer value representing the number of characters in this string.
Substrings. When you get a substring from a string—for example, using a subscript or a method like prefix(_:) —the result is an instance of Substring , not another string. Substrings in Swift have most of the same methods as strings, which means you can work with substrings the same way you work with strings.
How to Find SubString in Swift? To find substring of a String in Swift, prepare Range object using start and end indices, then give this Range object to the string in square brackets as an index. The syntax to concatenate two strings is:
Swift Substring. To find substring of a String in Swift, prepare range using start and end indexes, then use the range on the string. The syntax to concatenate two strings is: where startPosition and endPosition are integers that define the bounds of the substring in the main string str.
String slicing is somewhat verbose in Swift. If you’re coming from another language, it’s not immediately obvious how to do it at all. Many of us end up cursing the screen after trying to get substrings using patterns from other languages.
Swift Strings use a non-integer index type for reasons of speed and safety: Speed: because an index access will always be a constant-time operation. Safety: because you’re guaranteed to get a whole character and not, say, half of a 16-bit character. I know what you’re thinking: that’s swell and all, but I just want to grab a range of characters.
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.
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
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