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)
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"
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())
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.
Swift 4/5
var str = "bla"
str.removeLast() // returns "a"; str is now "bl"
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
Use the function removeAtIndex(i: String.Index) -> Character:
var s = "abc"
s.removeAtIndex(s.endIndex.predecessor()) // "ab"
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())
The easiest way to trim the last character of the string is:
title = title[title.startIndex ..< title.endIndex.advancedBy(-1)]
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
let str = "abc"
let substr = str.substringToIndex(str.endIndex.predecessor()) // "ab"
var str = "Hello, playground"
extension String {
var stringByDeletingLastCharacter: String {
return dropLast(self)
}
}
println(str.stringByDeletingLastCharacter) // "Hello, playgroun"
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.
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
Use the function advance(startIndex, endIndex):
var str = "45+22"
str = str.substringToIndex(advance(str.startIndex, countElements(str) - 1))
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"
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.
Swift 3 (according to the docs) 20th Nov 2016
let range = expression.index(expression.endIndex, offsetBy: -numberOfCharactersToRemove)..<expression.endIndex
expression.removeSubrange(range)
The dropLast() function removes the last element of the string.
var expression = "45+22"
expression = expression.dropLast()
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
}
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)
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
}
}
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.
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>
Related
in advance thanks for help.
I am trying to make calculator application (for specific purposes) and I would like to know, if there exist a way how to convert Double() to NSMutableAttributedString. I need this for label output answer.
Reason of using NSMutableAttributedString is because I would like to have answer with subscripts and upper-scripts.
//example of my code
var a = Double(), b = Double(), c = Double()
a = Double(textField1.text!)
b = Double(textField2.text!)
c = a + b
let font:UIFont? = UIFont(name: "Courier", size:12)
let fontSuper:UIFont? = UIFont(name: "Courier", size:10)
//for x_1 (subscript for "1")
x1_t:NSMutableAttributedString = NSMutableAttributedString(string: "x1", attributes: [NSFontAttributeName:font!])
x1_t.setAttributes([NSFontAttributeName:fontSuper!,NSBaselineOffsetAttributeName:-4], range: NSRange(location:1,length:1))
var result = NSMutableAttributedText()
// what to do to get output for a label like "x_1 = String(c) m"
If there exist another way like append String() to NSAtributedString() - I am looking forward for answers.
As I understand it, your input strings (named "prestring1" and "afterstring1" in your own answer) could just be normal strings without attributes, because you only need the final result to be an attributed string.
This would drastically simplify your function, for example you could use string interpolation first and then only make an attributed string and move up (or down) the last part (or any part you want, I'm using an hardcoded range in my example but it's just an example).
Like:
let randomstring = "Random ="
let afterstring = "m2"
let result: Double = 42.1
func stringer (pre: String,
result: Double,
post: String) -> NSMutableAttributedString
{
let base = "\(pre) \(result) \(post)"
let mutable = NSMutableAttributedString(string: base)
mutable.addAttribute(NSBaselineOffsetAttributeName, value: 4,
range: NSRange(location: mutable.length - 2, length: 2))
return mutable
}
let attributedString = stringer(pre: randomstring, result: result, post: afterstring)
Gives:
I am still not quit sure how to do it, but I could create simple function, which is approximately doing what I need. Here I am sharing my answer in case someone has the same question, but in case someone knows better answer, share it with others :)
var randomstring = "Random ="
var prestring1 = NSMutableAttributedString(string: randomstring)
var afterstring1 = NSMutableAttributedString(string: "m2")
var result1 = Double()
result1 = 42.1
func stringer (prestring: NSMutableAttributedString, result: Double, afterstring: NSMutableAttributedString) -> NSMutableAttributedString {
var mutableatributedresult = NSMutableAttributedString(string: String(result))
var mutableaddition = NSMutableAttributedString(string: " ")
var alltext = NSMutableAttributedString()
alltext.append(prestring)
alltext.append(mutableaddition)
alltext.append(mutableatributedresult)
alltext.append(mutableaddition)
alltext.append(afterstring)
return alltext
}
stringer(prestring: prestring1, result: result1, afterstring: afterstring1)
//This should give result of "Random = 42.1 m2"
If someone knows better solution I am curious.
A function in swift takes any numeric type in Swift (Int, Double, Float, UInt, etc).
the function converts the number to a string
the function signature is as follows :
func swiftNumbers <T : NumericType> (number : T) -> String {
//body
}
NumericType is a custom protocol that has been added to numeric types in Swift.
inside the body of the function, the number should be converted to a string:
I use the following
var stringFromNumber = "\(number)"
which is not so elegant, PLUS : if the absolute value of the number is strictly inferior to 0.0001 it gives this:
"\(0.000099)" //"9.9e-05"
or if the number is a big number :
"\(999999999999999999.9999)" //"1e+18"
is there a way to work around this string interpolation limitation? (without using Objective-C)
P.S :
NumberFormater doesn't work either
import Foundation
let number : NSNumber = 9_999_999_999_999_997
let formatter = NumberFormatter()
formatter.minimumFractionDigits = 20
formatter.minimumIntegerDigits = 20
formatter.minimumSignificantDigits = 40
formatter.string(from: number) // "9999999999999996.000000000000000000000000"
let stringFromNumber = String(format: "%20.20f", number) // "0.00000000000000000000"
Swift String Interpolation
1) Adding different types to a string
2) Means the string is created from a mix of constants, variables, literals or expressions.
Example:
let length:Float = 3.14
var breadth = 10
var myString = "Area of a rectangle is length*breadth"
myString = "\(myString) i.e. = \(length)*\(breadth)"
Output:
3.14
10
Area of a rectangle is length*breadth
Area of a rectangle is length*breadth i.e. = 3.14*10
Use the Swift String initializer: String(format: <#String#>, arguments: <#[CVarArgType]#>)
For example:
let stringFromNumber = String(format: "%.2f", number)
String and Characters conforms to StringInterpolationProtocol protocol which provide more power to the strings.
StringInterpolationProtocol - "Represents the contents of a string literal with interpolations while it’s being built up."
String interpolation has been around since the earliest days of Swift, but in Swift 5.0 it’s getting a massive overhaul to make it faster and more powerful.
let name = "Ashwinee Dhakde"
print("Hello, I'm \(name)")
Using the new string interpolation system in Swift 5.0 we can extend String.StringInterpolation to add our own custom interpolations, like this:
extension String.StringInterpolation {
mutating func appendInterpolation(_ value: Date) {
let formatter = DateFormatter()
formatter.dateStyle = .full
let dateString = formatter.string(from: value)
appendLiteral(dateString)
}
}
Usage: print("Today's date is \(Date()).")
We can even provide user-defined names to use String-Interpolation, let's understand with an example.
extension String.StringInterpolation {
mutating func appendInterpolation(JSON JSONData: Data) {
guard
let JSONObject = try? JSONSerialization.jsonObject(with: JSONData, options: []),
let jsonData = try? JSONSerialization.data(withJSONObject: JSONObject, options: .prettyPrinted) else {
appendInterpolation("Invalid JSON data")
return
}
appendInterpolation("\n\(String(decoding: jsonData, as: UTF8.self))")
}
}
print("The JSON is \(JSON: jsonData)")
Whenever we want to provide "JSON" in the string interpolation statement, it will print the .prettyPrinted
Isn't it cool!!
That is what i am trying to do:
var i = 0
var string = "abcdef"
for value in string
{
value.[Put value of variable i here] = "a"
i++
}
How can i insert the value of i in the code?
Easiest is probably just convert it to an NSMutableString:
let string = "abcdef".mutableCopy() as NSMutableString
println( "\(string)")
for var i = 0; i < string.length; ++i {
string.replaceCharactersInRange(NSMakeRange(i, 1), withString: "a")
}
println( "\(string)")
Yes, it's a bit ugly but it works.
A much cleaner way is to use Swifts map function:
var string = "abcdef"
let result = map(string) { (c) -> Character in
"a"
}
println("\(result)") // aaaaaa
You should just be able to use the following but this doesn't compile:
map(string) { "a" }
In you comments you mention you want to split up the string on a space, you can just use this for that:
let stringWithSpace = "abcdef 012345"
let splitString = stringWithSpace.componentsSeparatedByString(" ")
println("\(splitString[0])") // abcdef
println("\(splitString[1])") // 012345
How can I define a Dictionary with string key in AS3? and how to do read operation?
for example:
var Dic:Dictionary = new Dictionary();
Dic["Exhausted"] = "He who talks more is sooner exhausted, please keep smiling :)";
String str = str.substring(8,str.length-1); // == str = "Exhausted";
trace('Dic[' + str + '] = ' + Dic[str]);
the output is Dic[Exhausted] = undefined???!!
why?
I think you have a syntax error on this line:
String str = str.substring(8,str.length-1);
If you just use this :
var d:Dictionary = new Dictionary();
d["Exhausted"] = "He who talks more is sooner exhausted, please keep smiling :)";
trace(d["Exhausted"]);
You can see it's all fine.
The problem with var str:String = str.substring(8,str.length-1); is that you define a String named "str": var str:String, but you assign a value which is the result of the substring() method called on str and of cours str does not exist yet when you call substring on it.
Not sure if this makes sense: you define str as the result of applying substring on str.
The actionscript compiler should've complained btw:
expected a definition keyword (such as function) after attribute String, not str.
Just a wild guess, but would you happen to have str defined somewhere else in your code and you just update str again using str before using it ? In which case, you shouldn't redefine, str, you should simply assign a new value.
e.g.
//somewhere else
var str:String = "12345678Exhausted";
//further down
var Dic:Dictionary = new Dictionary();
Dic["Exhausted"] = "He who talks more is sooner exhausted, please keep smiling :)";
str = str.substring(8,str.length); // == str = "Exhausted";
trace('Dic[' + str + '] = ' + Dic[str]);
Also, another syntax error is how you define str:
should be var str:String (as2/as3 syntax), not String str(java/c++/etc. style)
What would be the best way to simply take a string like
var myString:String = "Thi$ i$ a T#%%Ible Exam73#";
and make myString = "thiiatibleeam";
or another example
var myString:String = "Totally Awesome String";
and make myString = "totallyawesomestring";
In actionscript 3 Thanks!
Extending #Sam OverMars' answer, you can use a combination of String's replace method with a Regex and String's toLowerCase method to get what you're looking for.
var str:String = "Thi$ i$ a T#%%Ible Exam73#";
str = str.toLowerCase(); //thi$ i$ a t#%%ible exam73#
str = str.replace(/[^a-z]/g,""); //thiiatibleexam
The regular expression means:
[^a-z] -- any character *not* in the range a-z
/g -- global tag means find all, not just find one
I think this is the regex you're looking for:
[Bindable]
var myString:String = "Thi$ i$ a T#%%Ible Exam73#";
[Bindable]
var anotherString:String = "";
protected function someFunction():void
{
anotherString = myString.replace(/[^a-zA-Z]/g, "");
anotherString = anotherString.toLowerCase();
}
I belive what your looking for is:
var myString = str.replace("find", "replace");
or in your case:
str.replace("$", "");
also, it might be:
str.replace('$', ' ');
//EDIT
How about:
var mySearch:RegExp = /(\t|\n|\s{1,})/g;
var myString = str.replace(mySearch, "");