How do you use String.substringWithRange? (or, how do Ranges work in Swift?)

For example to find the first name (up to the first space) in my full name:

let name = "Joris Kluivers"

let start = name.startIndex
let end = find(name, " ")

if end {
    let firstName = name[start..end!]
} else {
    // no space found
}

start and end are of type String.Index here and are used to create a Range<String.Index> and used in the subscript accessor (if a space is found at all in the original string).

It's hard to create a String.Index directly from an integer position as used in the opening post. This is because in my name each character would be of equal size in bytes. But characters using special accents in other languages could have used several more bytes (depending on the encoding used). So what byte should the integer refer to?

It's possible to create a new String.Index from an existing one using the methods succ and pred which will make sure the correct number of bytes are skipped to get to the next code point in the encoding. However in this case it's easier to search for the index of the first space in the string to find the end index.

---

Since String is a bridge type for NSString, the "old" methods should work, but it's not clear how - e.g., this doesn't work either (doesn't appear to be valid syntax): let x = str.substringWithRange(NSMakeRange(0, 3))

To me, that is the really interesting part of your question. String is bridged to NSString, so most NSString methods do work directly on a String. You can use them freely and without thinking. So, for example, this works just as you expect:

// delete all spaces from Swift String stateName
stateName = stateName.stringByReplacingOccurrencesOfString(" ", withString:"")

But, as so often happens, "I got my mojo workin' but it just don't work on you." You just happened to pick one of the rare cases where a parallel identically named Swift method exists, and in a case like that, the Swift method overshadows the Objective-C method. Thus, when you say str.substringWithRange, Swift thinks you mean the Swift method rather than the NSString method — and then you are hosed, because the Swift method expects a Range<String.Index>, and you don't know how to make one of those.

The easy way out is to stop Swift from overshadowing like this, by casting explicitly:

let x = (str as NSString).substringWithRange(NSMakeRange(0, 3))

Note that no significant extra work is involved here. "Cast" does not mean "convert"; the String is effectively an NSString. We are just telling Swift how to look at this variable for purposes of this one line of code.

The really weird part of this whole thing is that the Swift method, which causes all this trouble, is undocumented. I have no idea where it is defined; it is not in the NSString header and it's not in the Swift header either.

---

The short answer is that this is really hard in Swift right now. My hunch is that there is still a bunch of work for Apple to do on convenience methods for things like this.

String.substringWithRange() is expecting a Range<String.Index> parameter, and as far as I can tell there isn't a generator method for the String.Index type. You can get String.Index values back from aString.startIndex and aString.endIndex and call .succ() or .pred() on them, but that's madness.

How about an extension on the String class that takes good old Ints?

extension String {
    subscript (r: Range<Int>) -> String {
        get {
            let subStart = advance(self.startIndex, r.startIndex, self.endIndex)
            let subEnd = advance(subStart, r.endIndex - r.startIndex, self.endIndex)
            return self.substringWithRange(Range(start: subStart, end: subEnd))
        }
    }
    func substring(from: Int) -> String {
        let end = countElements(self)
        return self[from..<end]
    }
    func substring(from: Int, length: Int) -> String {
        let end = from + length
        return self[from..<end]
    }
}

let mobyDick = "Call me Ishmael."
println(mobyDick[8...14])             // Ishmael

let dogString = "This 🐶's name is Patch."
println(dogString[5..<6])               // 🐶
println(dogString[5...5])              // 🐶
println(dogString.substring(5))        // 🐶's name is Patch.
println(dogString.substring(5, length: 1))   // 🐶

Update: Swift beta 4 resolves the issues below!

As it stands [in beta 3 and earlier], even Swift-native strings have some issues with handling Unicode characters. The dog icon above worked, but the following doesn't:

let harderString = "1:1️⃣"
for character in harderString {
    println(character)
}

Output:

1
:
1
️
⃣

---

In new Xcode 7.0 use

//: Playground - noun: a place where people can play

import UIKit

var name = "How do you use String.substringWithRange?"
let range = name.startIndex.advancedBy(0)..<name.startIndex.advancedBy(10)
name.substringWithRange(range)

//OUT:

https://i.stack.imgur.com/1wYAl.png

---

You can use this extensions to improve substringWithRange

Swift 2.3

extension String
{   
    func substringWithRange(start: Int, end: Int) -> String
    {
        if (start < 0 || start > self.characters.count)
        {
            print("start index \(start) out of bounds")
            return ""
        }
        else if end < 0 || end > self.characters.count
        {
            print("end index \(end) out of bounds")
            return ""
        }
        let range = Range(start: self.startIndex.advancedBy(start), end: self.startIndex.advancedBy(end))
        return self.substringWithRange(range)
    }

    func substringWithRange(start: Int, location: Int) -> String
    {
        if (start < 0 || start > self.characters.count)
        {
            print("start index \(start) out of bounds")
            return ""
        }
        else if location < 0 || start + location > self.characters.count
        {
            print("end index \(start + location) out of bounds")
            return ""
        }
        let range = Range(start: self.startIndex.advancedBy(start), end: self.startIndex.advancedBy(start + location))
        return self.substringWithRange(range)
    }
}

Swift 3

extension String
{
    func substring(start: Int, end: Int) -> String
    {
        if (start < 0 || start > self.characters.count)
        {
            print("start index \(start) out of bounds")
            return ""
        }
        else if end < 0 || end > self.characters.count
        {
            print("end index \(end) out of bounds")
            return ""
        }
        let startIndex = self.characters.index(self.startIndex, offsetBy: start)
        let endIndex = self.characters.index(self.startIndex, offsetBy: end)
        let range = startIndex..<endIndex

        return self.substring(with: range)
    }

    func substring(start: Int, location: Int) -> String
    {
        if (start < 0 || start > self.characters.count)
        {
            print("start index \(start) out of bounds")
            return ""
        }
        else if location < 0 || start + location > self.characters.count
        {
            print("end index \(start + location) out of bounds")
            return ""
        }
        let startIndex = self.characters.index(self.startIndex, offsetBy: start)
        let endIndex = self.characters.index(self.startIndex, offsetBy: start + location)
        let range = startIndex..<endIndex

        return self.substring(with: range)
    }
}

Usage:

let str = "Hello, playground"

let substring1 = str.substringWithRange(0, end: 5) //Hello
let substring2 = str.substringWithRange(7, location: 10) //playground

---

Sample Code for how to get substring in Swift 2.0

(i) Substring from starting index

Input:-

var str = "Swift is very powerful language!"
print(str)

str = str.substringToIndex(str.startIndex.advancedBy(5))
print(str)

Output:-

Swift is very powerful language!
Swift

(ii) Substring from particular index

Input:-

var str = "Swift is very powerful language!"
print(str)

str = str.substringFromIndex(str.startIndex.advancedBy(6)).substringToIndex(str.startIndex.advancedBy(2))
print(str)

Output:-

Swift is very powerful language!
is

I hope it will help you!

---

Easy solution with little code.

Make an extension that includes basic subStringing that nearly all other languages have:

extension String {
    func subString(start: Int, end: Int) -> String {
        let startIndex = self.index(self.startIndex, offsetBy: start)
        let endIndex = self.index(startIndex, offsetBy: end)

        let finalString = self.substring(from: startIndex)
        return finalString.substring(to: endIndex)
    }
}

Simply call this with

someString.subString(start: 0, end: 6)

---

This works in my playground :)

String(seq: Array(str)[2...4])

---

Updated for Xcode 7. Adds String extension:

Use:

var chuck: String = "Hello Chuck Norris"
chuck[6...11] // => Chuck

Implementation:

extension String {

    /**
     Subscript to allow for quick String substrings ["Hello"][0...1] = "He"
     */
    subscript (r: Range<Int>) -> String {
        get {
            let start = self.startIndex.advancedBy(r.startIndex)
            let end = self.startIndex.advancedBy(r.endIndex - 1)
            return self.substringWithRange(start..<end)
        }
    }

}

---

try this in playground

var str:String = "Hello, playground"

let range = Range(start:advance(str.startIndex,1), end: advance(str.startIndex,8))

it will give you "ello, p"

However where this gets really interesting is that if you make the last index bigger than the string in playground it will show any strings that you defined after str :o

Range() appears to be a generic function so that it needs to know the type it is dealing with.

You also have to give it the actual string your interested in playgrounds as it seems to hold all stings in a sequence one after another with their variable name afterwards.

So

var str:String = "Hello, playground"

var str2:String = "I'm the next string"

let range = Range(start:advance(str.startIndex,1), end: advance(str.startIndex,49))

gives "ello, playground�str���I'm the next string�str2�"

works even if str2 is defined with a let

:)

---

Rob Napier had already given a awesome answer using subscript. But i felt one drawback in that as there is no check for out of bound conditions. This can tend to crash. So i modified the extension and here it is

extension String {
    subscript (r: Range<Int>) -> String? { //Optional String as return value
        get {
            let stringCount = self.characters.count as Int
            //Check for out of boundary condition
            if (stringCount < r.endIndex) || (stringCount < r.startIndex){
                return nil
            }
            let startIndex = self.startIndex.advancedBy(r.startIndex)

            let endIndex = self.startIndex.advancedBy(r.endIndex - r.startIndex)

            return self[Range(start: startIndex, end: endIndex)]
        }
    }
}

Output below

var str2 = "Hello, World"

var str3 = str2[0...5]
//Hello,
var str4 = str2[0..<5]
//Hello
var str5 = str2[0..<15]
//nil

So i suggest always to check for the if let

if let string = str[0...5]
{
    //Manipulate your string safely
}

---

In Swift3

For ex: a variable "Duke James Thomas", we need to get "James".

let name = "Duke James Thomas"
let range: Range<String.Index> = name.range(of:"James")!
let lastrange: Range<String.Index> = img.range(of:"Thomas")!
var middlename = name[range.lowerBound..<lstrange.lowerBound]
print (middlename)

---

Taking a page from Rob Napier, I developed these Common String Extensions, two of which are:

subscript (r: Range<Int>) -> String
{
    get {
        let startIndex = advance(self.startIndex, r.startIndex)
        let endIndex = advance(self.startIndex, r.endIndex - 1)

        return self[Range(start: startIndex, end: endIndex)]
    }
}

func subString(startIndex: Int, length: Int) -> String
{
    var start = advance(self.startIndex, startIndex)
    var end = advance(self.startIndex, startIndex + length)
    return self.substringWithRange(Range<String.Index>(start: start, end: end))
}

Usage:

"Awesome"[3...7] //"some"
"Awesome".subString(3, length: 4) //"some"

---

This is how you get a range from a string:

var str = "Hello, playground"

let startIndex = advance(str.startIndex, 1)
let endIndex = advance(startIndex, 8)
let range = startIndex..<endIndex
let substr = str[range] //"ello, pl"

The key point is that you are passing a range of values of type String.Index (this is what advance returns) instead of integers.

The reason why this is necessary, is that strings in Swift don't have random access (because of variable length of Unicode characters basically). You also can't do str[1]. String.Index is designed to work with their internal structure.

You can create an extension with a subscript though, that does this for you, so you can just pass a range of integers (see e.g. Rob Napier's answer).

---

I tried to come up with something Pythonic.

All the subscripts here are great, but the times I really need something simple is usually when I want to count from back, e.g. string.endIndex.advancedBy(-1)

It supports nil values, for both start and end index, where nil would mean index at 0 for start, string.characters.count for end.

extension String {
    var subString: (Int?) -> (Int?) -> String {
        return { (start) in
            { (end) in
                let startIndex = start ?? 0 < 0 ? self.endIndex.advancedBy(start!) : self.startIndex.advancedBy(start ?? 0)
                let endIndex = end ?? self.characters.count < 0 ? self.endIndex.advancedBy(end!) : self.startIndex.advancedBy(end ?? self.characters.count)

                return startIndex > endIndex ? "" : self.substringWithRange(startIndex ..< endIndex)
            }
        }
    }
}

let dog = "Dog‼🐶"
print(dog.subString(nil)(-1)) // Dog!!

EDIT

A more elegant solution:

public extension String {
    struct Substring {
        var start: Int?
        var string: String

        public subscript(end: Int?) -> String {
            let startIndex = start ?? 0 < 0 ? string.endIndex.advancedBy(start!) : string.startIndex.advancedBy(start ?? 0)
            let endIndex = end ?? string.characters.count < 0 ? string.endIndex.advancedBy(end!) : string.startIndex.advancedBy(end ?? string.characters.count)

            return startIndex > endIndex ? "" : string.substringWithRange(startIndex ..< endIndex)
        }
    }

    public subscript(start: Int?) -> Substring {
        return Substring(start: start, string: self)
    }
}

let dog = "Dog‼🐶"
print(dog[nil][-1]) // Dog!!

---

First create the range, then the substring. You can use fromIndex..<toIndex syntax like so:

let range = fullString.startIndex..<fullString.startIndex.advancedBy(15) // 15 first characters of the string
let substring = fullString.substringWithRange(range)

---

here is a example to get video-Id only .i.e (6oL687G0Iso) from the whole URL in swift

let str = "https://www.youtube.com/watch?v=6oL687G0Iso&list=PLKmzL8Ib1gsT-5LN3V2h2H14wyBZTyvVL&index=2"
var arrSaprate = str.componentsSeparatedByString("v=")
let start = arrSaprate[1]
let rangeOfID = Range(start: start.startIndex,end:start.startIndex.advancedBy(11))
let substring = start[rangeOfID]
print(substring)

---

let startIndex = text.startIndex
var range = startIndex.advancedBy(1) ..< text.endIndex.advancedBy(-4)
let substring = text.substringWithRange(range)

Full sample you can see here

---

http://www.learnswiftonline.com/reference-guides/string-reference-guide-for-swift/ shows that this works well:

var str = "abcd"
str = str.substringToIndex(1)

---

Well, I had the same issue and solved with the "bridgeToObjectiveC()" function:

var helloworld = "Hello World!"
var world = helloworld.bridgeToObjectiveC().substringWithRange(NSMakeRange(6,6))
println("\(world)") // should print World!

Please note that in the example, substringWithRange in conjunction with NSMakeRange take the part of the string starting at index 6 (character "W") and finishing at index 6 + 6 positions ahead (character "!")

Cheers.

---

You can use any of the substring methods in a Swift String extension I wrote https://bit.ly/JString.

var string = "hello"
var sub = string.substringFrom(3) // or string[3...5]
println(sub)// "lo"

---

If you have an NSRange, bridging to NSString works seamlessly. For example, I was doing some work with UITextFieldDelegate and I quickly wanted to compute the new string value when it asked if it should replace the range.

func textField(textField: UITextField, shouldChangeCharactersInRange range: NSRange, replacementString string: String) -> Bool {
    let newString = (textField.text as NSString).stringByReplacingCharactersInRange(range, withString: string)
    println("Got new string: ", newString)
}

---

Simple extension for String:

extension String {

    func substringToIndex(index: Int) -> String {
        return self[startIndex...startIndex.advancedBy(min(index, characters.count - 1))]
    }
}

---

If you don't care about performance... this is probably the most concise solution in Swift 4

extension String {
    subscript(range: CountableClosedRange<Int>) -> String {
        return enumerated().filter{$0.offset >= range.first! && $0.offset < range.last!}
            .reduce(""){$0 + String($1.element)}
    }
}

It enables you to do something like this:

let myStr = "abcd"
myStr[0..<2] // produces "ab"

---