Remove last character from string. Swift language

Question

How can I remove last character from String variable using Swift? Can't find it in documentation.

Here is full example:

var expression = "45+22"
expression = expression.substringToIndex(countElements(expression) - 1)

Andromache

2017, MODERN ANSWER: https://mcmap.net/q/23518/-delete-last-character-in-swift-3 – Ilion

Answer 1

Swift 4.0 (also Swift 5.0)

var str = "Hello, World"                           // "Hello, World"
str.dropLast()                                     // "Hello, Worl" (non-modifying)
str                                                // "Hello, World"
String(str.dropLast())                             // "Hello, Worl"

str.remove(at: str.index(before: str.endIndex))    // "d"
str                                                // "Hello, Worl" (modifying)

Swift 3.0

The APIs have gotten a bit more swifty, and as a result the Foundation extension has changed a bit:

var name: String = "Dolphin"
var truncated = name.substring(to: name.index(before: name.endIndex))
print(name)      // "Dolphin"
print(truncated) // "Dolphi"

Or the in-place version:

var name: String = "Dolphin"
name.remove(at: name.index(before: name.endIndex))
print(name)      // "Dolphi"

Thanks Zmey, Rob Allen!

Swift 2.0+ Way

There are a few ways to accomplish this:

Via the Foundation extension, despite not being part of the Swift library:

var name: String = "Dolphin"
var truncated = name.substringToIndex(name.endIndex.predecessor())
print(name)      // "Dolphin"
print(truncated) // "Dolphi"

Using the removeRange() method (which alters the name):

var name: String = "Dolphin"    
name.removeAtIndex(name.endIndex.predecessor())
print(name) // "Dolphi"

Using the dropLast() function:

var name: String = "Dolphin"
var truncated = String(name.characters.dropLast())
print(name)      // "Dolphin"
print(truncated) // "Dolphi"

Old String.Index (Xcode 6 Beta 4 +) Way

Since String types in Swift aim to provide excellent UTF-8 support, you can no longer access character indexes/ranges/substrings using Int types. Instead, you use String.Index:

let name: String = "Dolphin"
let stringLength = count(name) // Since swift1.2 `countElements` became `count`
let substringIndex = stringLength - 1
name.substringToIndex(advance(name.startIndex, substringIndex)) // "Dolphi"

Alternatively (for a more practical, but less educational example) you can use endIndex:

let name: String = "Dolphin"
name.substringToIndex(name.endIndex.predecessor()) // "Dolphi"

Note: I found this to be a great starting point for understanding String.Index

Old (pre-Beta 4) Way

You can simply use the substringToIndex() function, providing it one less than the length of the String:

let name: String = "Dolphin"
name.substringToIndex(countElements(name) - 1) // "Dolphi"

Glamorous

On Xcode 6 beta 6 :'String' doesn't have a member named substringToIndex – Cline

@Carpsen90 There is no beta 6. (As of this writing.) And in the most recent beta (beta 4), the String type definitely has a substringToIndex() method. Can you clarify? – Glamorous

I am sorry I meant beta 4.. and it is definitely not working for me.. I had to cast the string to NSString to have access to more functions – Cline

@Carpsen90 Can you maybe ask as a separate question? – Glamorous

Hey, heads up, make sure it's substringToIndex not substringFromIndex. It doesn't make you feel intelligent mistaking this, let me tell you. – Nkvd

I just renamed countElements to count in the Beta4+ way part of the answer. (Swift1.2) – Applewhite

This is true, as of July 21st, Xcode 7 Beta 4 says 'String' doesn't have a member named substringToIndex. Also, as of Xcode 7, string no longer has a .count property, it is now only applied to characters: string.characters.count – Moulin

looks like advance doesn't work anymore either in Xcode 7 beta 6, do you know of a work around? – Reinforcement

Swift 3: var truncated = name.substring(to: name.index(before: name.endIndex)) – Franchescafranchise

Swift 3 in place: name.remove(at: name.index(before: name.endIndex)) – Raleigh

Your answer is the history of Swift evolution. – Nonentity

Note that in Swift 4 strings are back to being collections, and the .substring(to:) method is deprecated. The options listed under Swift 2.0+ worked for me in Swift 4. – Kamalakamaria

That's not working in Swift 4, Xcode 9. So I did this placeStr.removeLast(1) which means remove the last character from the string and returns Void, and it did well. – Determinate

var truncated = name.substring(to: name.index(before: name.endIndex)) This worked for me in Swift 3 Thanks @CraigOtis – Gobetween

Answer 2

The global dropLast() function works on sequences and therefore on Strings:

var expression  = "45+22"
expression = dropLast(expression)  // "45+2"

// in Swift 2.0 (according to cromanelli's comment below)
expression = String(expression.characters.dropLast())

Tabaret

In Swift 2.0 the characters property on a String outputs a sequence, therefore now you have to use: expression = expression.characters.dropLast() – Tabaret

In Swift 2.0 for have to properly cast the result expression = String(expression.characters.dropLast()) if you want it back as a String – Treehopper

Answer 3

Swift 4:

let choppedString = String(theString.dropLast())

In Swift 2, do this:

let choppedString = String(theString.characters.dropLast())

I recommend this link to get an understanding of Swift strings.

Virgy

Could be just let choppedString = theString.dropLast().description. – Ectoplasm

Answer 4

Swift 4/5

var str = "bla"
str.removeLast() // returns "a"; str is now "bl"

Salts

Answer 5

This is a String Extension Form:

extension String {

    func removeCharsFromEnd(count_:Int) -> String {
        let stringLength = count(self)

        let substringIndex = (stringLength < count_) ? 0 : stringLength - count_

        return self.substringToIndex(advance(self.startIndex, substringIndex))
    }
}

for versions of Swift earlier than 1.2:

...
let stringLength = countElements(self)
...

---

Usage:

var str_1 = "Maxim"
println("output: \(str_1.removeCharsFromEnd(1))") // "Maxi"
println("output: \(str_1.removeCharsFromEnd(3))") // "Ma"
println("output: \(str_1.removeCharsFromEnd(8))") // ""

---

Reference:

Extensions add new functionality to an existing class, structure, or enumeration type. This includes the ability to extend types for which you do not have access to the original source code (known as retroactive modeling). Extensions are similar to categories in Objective-C. (Unlike Objective-C categories, Swift extensions do not have names.)

See DOCS

Astred

Hi, have you test emojis? – Garda

@ZuYuan what is emojis? – Astred

Cannot find 'advance' in scope – Cowley

Answer 6

Use the function removeAtIndex(i: String.Index) -> Character:

var s = "abc"    
s.removeAtIndex(s.endIndex.predecessor())  // "ab"

Eclampsia

Answer 7

Swift 4

var welcome = "Hello World!"
welcome = String(welcome[..<welcome.index(before:welcome.endIndex)])

or

welcome.remove(at: welcome.index(before: welcome.endIndex))

or

welcome = String(welcome.dropLast())

Determinism

Answer 8

The easiest way to trim the last character of the string is:

title = title[title.startIndex ..< title.endIndex.advancedBy(-1)]

Demicanton

Thank you, a very elegant way to remove as many characters as you want from the end of a string. – Alidus

Answer 9

import UIKit

var str1 = "Hello, playground"
str1.removeLast()
print(str1)

var str2 = "Hello, playground"
str2.removeLast(3)
print(str2)

var str3 = "Hello, playground"
str3.removeFirst(2)
print(str3)

Output:-
Hello, playgroun
Hello, playgro
llo, playground

Fetich

Answer 10

let str = "abc"
let substr = str.substringToIndex(str.endIndex.predecessor())  // "ab"

Wardle

Answer 11

var str = "Hello, playground"

extension String {
    var stringByDeletingLastCharacter: String {
        return dropLast(self)
    }
}

println(str.stringByDeletingLastCharacter)   // "Hello, playgroun"

Gorget

kkkk I didn't even remember of this one (Swift 1.2 above). for Swift 3 version https://mcmap.net/q/23520/-remove-last-two-characters-in-a-string – Gorget

Answer 12

Short answer (valid as of 2015-04-16): removeAtIndex(myString.endIndex.predecessor())

Example:

var howToBeHappy = "Practice compassion, attention and gratitude. And smile!!"
howToBeHappy.removeAtIndex(howToBeHappy.endIndex.predecessor())
println(howToBeHappy)
// "Practice compassion, attention and gratitude. And smile!"

Meta:

The language continues its rapid evolution, making the half-life for many formerly-good S.O. answers dangerously brief. It's always best to learn the language and refer to real documentation.

Everyday

Answer 13

With the new Substring type usage:

Swift 4:

var before: String = "Hello world!"
var lastCharIndex: Int = before.endIndex
var after:String = String(before[..<lastCharIndex])
print(after) // Hello world

Shorter way:

var before: String = "Hello world!"
after = String(before[..<before.endIndex])
print(after) // Hello world

Hiller

Answer 14

Use the function advance(startIndex, endIndex):

var str = "45+22"
str = str.substringToIndex(advance(str.startIndex, countElements(str) - 1))

Sewn

Answer 15

A swift category that's mutating:

extension String {
    mutating func removeCharsFromEnd(removeCount:Int)
    {
        let stringLength = count(self)
        let substringIndex = max(0, stringLength - removeCount)
        self = self.substringToIndex(advance(self.startIndex, substringIndex))
    }
}

Use:

var myString = "abcd"
myString.removeCharsFromEnd(2)
println(myString) // "ab"

Horselaugh

Answer 16

Another way If you want to remove one or more than one character from the end.

var myStr = "Hello World!"
myStr = (myStr as NSString).substringToIndex((myStr as NSString).length-XX)

Where XX is the number of characters you want to remove.

Desiredesirea

Answer 17

Swift 3 (according to the docs) 20th Nov 2016

let range = expression.index(expression.endIndex, offsetBy: -numberOfCharactersToRemove)..<expression.endIndex
expression.removeSubrange(range)

Warmup

Answer 18

The dropLast() function removes the last element of the string.

var expression = "45+22"
expression = expression.dropLast()

Bihar

Answer 19

Swift 4.2

I also delete my last character from String (i.e. UILabel text) in IOS app

@IBOutlet weak var labelText: UILabel! // Do Connection with UILabel

@IBAction func whenXButtonPress(_ sender: UIButton) { // Do Connection With X Button

    labelText.text = String((labelText.text?.dropLast())!) // Delete the last caracter and assign it

}

Jozef

Answer 20

I'd recommend using NSString for strings that you want to manipulate. Actually come to think of it as a developer I've never run into a problem with NSString that Swift String would solve... I understand the subtleties. But I've yet to have an actual need for them.

var foo = someSwiftString as NSString

or

var foo = "Foo" as NSString

or

var foo: NSString = "blah"

And then the whole world of simple NSString string operations is open to you.

As answer to the question

// check bounds before you do this, e.g. foo.length > 0
// Note shortFoo is of type NSString
var shortFoo = foo.substringToIndex(foo.length-1)

Habitude

Answer 21

complimentary to the above code I wanted to remove the beginning of the string and could not find a reference anywhere. Here is how I did it:

var mac = peripheral.identifier.description
let range = mac.startIndex..<mac.endIndex.advancedBy(-50)
mac.removeRange(range)  // trim 17 characters from the beginning
let txPower = peripheral.advertisements.txPower?.description

This trims 17 characters from the beginning of the string (he total string length is 67 we advance -50 from the end and there you have it.

Clavier

Answer 22

Swift 3: When you want to remove trailing string:

func replaceSuffix(_ suffix: String, replacement: String) -> String {
    if hasSuffix(suffix) {
        let sufsize = suffix.count < count ? -suffix.count : 0
        let toIndex = index(endIndex, offsetBy: sufsize)
        return substring(to: toIndex) + replacement
    }
    else
    {
        return self
    }
}

Ephemerality

Answer 23

I prefer the below implementation because I don't have to worry even if the string is empty

let str = "abc"
str.popLast()

// Prints ab

str = ""
str.popLast() // It returns the Character? which is an optional

// Print <emptystring>

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