How To Create Range in Swift?

Question

In Objective-c we create range by using NSRange

NSRange range;

So how to create range in Swift?

Answer 1

(1..<10)

returns...

Range = 1..<10

Answer 2

I created the following extension:

extension String {
    func substring(from from:Int, to:Int) -> String? {
        if from<to && from>=0 && to<self.characters.count {
            let rng = self.startIndex.advancedBy(from)..<self.startIndex.advancedBy(to)
            return self.substringWithRange(rng)
        } else {
            return nil
        }
    }
}

example of use:

print("abcde".substring(from: 1, to: 10)) //nil
print("abcde".substring(from: 2, to: 4))  //Optional("cd")
print("abcde".substring(from: 1, to: 0))  //nil
print("abcde".substring(from: 1, to: 1))  //nil
print("abcde".substring(from: -1, to: 1)) //nil

Answer 3

Use like this

var start = str.startIndex // Start at the string's start index
var end = advance(str.startIndex, 5) // Take start index and advance 5 characters forward
var range: Range<String.Index> = Range<String.Index>(start: start,end: end)

let firstFiveDigit =  str.substringWithRange(range)

print(firstFiveDigit)

Output : Hello

Answer 4

I find it surprising that, even in Swift 4, there's still no simple native way to express a String range using Int. The only String methods that let you supply an Int as a way of obtaining a substring by range are prefix and suffix.

It is useful to have on hand some conversion utilities, so that we can talk like NSRange when speaking to a String. Here's a utility that takes a location and length, just like NSRange, and returns a Range<String.Index>:

func range(_ start:Int, _ length:Int) -> Range<String.Index> {
    let i = self.index(start >= 0 ? self.startIndex : self.endIndex,
        offsetBy: start)
    let j = self.index(i, offsetBy: length)
    return i..<j
}

For example, "hello".range(0,1)" is the Range<String.Index> embracing the first character of "hello". As a bonus, I've allowed negative locations: "hello".range(-1,1)" is the Range<String.Index> embracing the last character of "hello".

It is useful also to convert a Range<String.Index> to an NSRange, for those moments when you have to talk to Cocoa (for example, in dealing with NSAttributedString attribute ranges). Swift 4 provides a native way to do that:

let nsrange = NSRange(range, in:s) // where s is the string

We can thus write another utility where we go directly from a String location and length to an NSRange:

extension String {
    func nsRange(_ start:Int, _ length:Int) -> NSRange {
        return NSRange(self.range(start,length), in:self)
    }
}

Answer 5

Xcode 8 beta 2 • Swift 3

let myString = "Hello World"
let myRange = myString.startIndex..<myString.index(myString.startIndex, offsetBy: 5)
let mySubString = myString.substring(with: myRange)   // Hello

---

Xcode 7 • Swift 2.0

let myString = "Hello World"
let myRange = Range<String.Index>(start: myString.startIndex, end: myString.startIndex.advancedBy(5))

let mySubString = myString.substringWithRange(myRange)   // Hello

or simply

let myString = "Hello World"
let myRange = myString.startIndex..<myString.startIndex.advancedBy(5)
let mySubString = myString.substringWithRange(myRange)   // Hello

Answer 6

If anyone want to create NSRange object can create as:

let range: NSRange = NSRange.init(location: 0, length: 5)

this will create range with position 0 and length 5