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.
Answer from user3151675 on Stack OverflowYou 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
You can get the ranges of your substrings using a while loop to repeat the search from that point to the end of your string and use map to get the substrings from the resulting ranges:
extension StringProtocol {
func ranges<S:StringProtocol,T:StringProtocol>(between start: S, and end: T, options: String.CompareOptions = []) -> [Range<Index>] {
var ranges: [Range<Index>] = []
var startIndex = self.startIndex
while startIndex < endIndex,
let lower = self[startIndex...].range(of: start, options: options)?.upperBound,
let range = self[lower...].range(of: end, options: options) {
let upper = range.lowerBound
ranges.append(lower..<upper)
startIndex = range.upperBound
}
return ranges
}
func substrings<S:StringProtocol,T:StringProtocol>(between start: S, and end: T, options: String.CompareOptions = []) -> [SubSequence] {
ranges(between: start, and: end, options: options).map{self[$0]}
}
}
Playground testing:
let string = """
your text
id:244476end
id:383848448end
id:55678900end
the end
"""
let substrings = string.substrings(between: "id:", and: "end") // ["244476", "383848448", "55678900"]
Rather thant trying to parse the string from start to end, I would use a combination of existing methods to transform it into the desire result. Here's How I would do this:
import Foundation
let raw = "id:244476end36475677id:383848448end334566777788id:55678900end543"
let result = raw
.components(separatedBy: "id:")
.filter{ !$0.isEmpty }
.map { segment -> String in
let slices = segment.components(separatedBy: "end")
return slices.first! // Removes the `end` and everything thereafter
}
print(result) // => ["244476", "383848448", "55678900"]
The problem is that mentions[0...3] returns an ArraySlice<String>, not an Array<String>. Therefore you could first use the Array(_:) initialiser in order to convert the slice into an array:
let first3Elements : [String] // An Array of up to the first 3 elements.
if mentions.count >= 3 {
first3Elements = Array(mentions[0 ..< 3])
} else {
first3Elements = mentions
}
Or if you want to use an ArraySlice (they are useful for intermediate computations, as they present a 'view' onto the original array, but are not designed for long term storage), you could subscript mentions with the full range of indices in your else:
let slice : ArraySlice<String> // An ArraySlice of up to the first 3 elements
if mentions.count >= 3 {
slice = mentions[0 ..< 3]
} else {
slice = mentions[mentions.indices] // in Swift 4: slice = mentions[...]
}
Although the simplest solution by far would be just to use the prefix(_:) method, which will return an ArraySlice of the first n elements, or a slice of the entire array if n exceeds the array count:
let slice = mentions.prefix(3) // ArraySlice of up to the first 3 elements
We can do like this,
let arr = [10,20,30,40,50]
let slicedArray = arr[1...3]
if you want to convert sliced array to normal array,
let arrayOfInts = Array(slicedArray)