Trim only trailing whitespace from end of string in Swift 3 - swift

Every example of trimming strings in Swift remove both leading and trailing whitespace, but how can only trailing whitespace be removed?
For example, if I have a string:
" example "
How can I end up with:
" example"
Every solution I've found shows trimmingCharacters(in: CharacterSet.whitespaces), but I want to retain the leading whitespace.
RegEx is a possibility, or a range can be derived to determine index of characters to remove, but I can't seem to find an elegant solution for this.

With regular expressions:
let string = " example "
let trimmed = string.replacingOccurrences(of: "\\s+$", with: "", options: .regularExpression)
print(">" + trimmed + "<")
// > example<
\s+ matches one or more whitespace characters, and $ matches
the end of the string.

In Swift 4 & Swift 5
This code will also remove trailing new lines.
It works based on a Character struct's method .isWhitespace
var trailingSpacesTrimmed: String {
var newString = self
while newString.last?.isWhitespace == true {
newString = String(newString.dropLast())
}
return newString
}

This short Swift 3 extension of string uses the .anchored and .backwards option of rangeOfCharacter and then calls itself recursively if it needs to loop. Because the compiler is expecting a CharacterSet as the parameter, you can just supply the static when calling, e.g. "1234 ".trailing(.whitespaces) will return "1234". (I've not done timings, but would expect faster than regex.)
extension String {
func trailingTrim(_ characterSet : CharacterSet) -> String {
if let range = rangeOfCharacter(from: characterSet, options: [.anchored, .backwards]) {
return self.substring(to: range.lowerBound).trailingTrim(characterSet)
}
return self
}
}

In Foundation you can get ranges of indices matching a regular expression. You can also replace subranges. Combining this, we get:
import Foundation
extension String {
func trimTrailingWhitespace() -> String {
if let trailingWs = self.range(of: "\\s+$", options: .regularExpression) {
return self.replacingCharacters(in: trailingWs, with: "")
} else {
return self
}
}
}
You can also have a mutating version of this:
import Foundation
extension String {
mutating func trimTrailingWhitespace() {
if let trailingWs = self.range(of: "\\s+$", options: .regularExpression) {
self.replaceSubrange(trailingWs, with: "")
}
}
}
If we match against \s* (as Martin R. did at first) we can skip the if let guard and force-unwrap the optional since there will always be a match. I think this is nicer since it's obviously safe, and remains safe if you change the regexp. I did not think about performance.

Handy String extension In Swift 4
extension String {
func trimmingTrailingSpaces() -> String {
var t = self
while t.hasSuffix(" ") {
t = "" + t.dropLast()
}
return t
}
mutating func trimmedTrailingSpaces() {
self = self.trimmingTrailingSpaces()
}
}

Swift 4
extension String {
var trimmingTrailingSpaces: String {
if let range = rangeOfCharacter(from: .whitespacesAndNewlines, options: [.anchored, .backwards]) {
return String(self[..<range.lowerBound]).trimmingTrailingSpaces
}
return self
}
}

Demosthese's answer is a useful solution to the problem, but it's not particularly efficient. This is an upgrade to their answer, extending StringProtocol instead, and utilizing Substring to remove the need for repeated copying.
extension StringProtocol {
#inline(__always)
var trailingSpacesTrimmed: Self.SubSequence {
var view = self[...]
while view.last?.isWhitespace == true {
view = view.dropLast()
}
return view
}
}

No need to create a new string when dropping from the end each time.
extension String {
func trimRight() -> String {
String(reversed().drop { $0.isWhitespace }.reversed())
}
}
This operates on the collection and only converts the result back into a string once.

It's a little bit hacky :D
let message = " example "
var trimmed = ("s" + message).trimmingCharacters(in: .whitespacesAndNewlines)
trimmed = trimmed.substring(from: trimmed.index(after: trimmed.startIndex))

Without regular expression there is not direct way to achieve that.Alternatively you can use the below function to achieve your required result :
func removeTrailingSpaces(with spaces : String) -> String{
var spaceCount = 0
for characters in spaces.characters{
if characters == " "{
print("Space Encountered")
spaceCount = spaceCount + 1
}else{
break;
}
}
var finalString = ""
let duplicateString = spaces.replacingOccurrences(of: " ", with: "")
while spaceCount != 0 {
finalString = finalString + " "
spaceCount = spaceCount - 1
}
return (finalString + duplicateString)
}
You can use this function by following way :-
let str = " Himanshu "
print(removeTrailingSpaces(with : str))

One line solution with Swift 4 & 5
As a beginner in Swift and iOS programming I really like #demosthese's solution above with the while loop as it's very easy to understand. However the example code seems longer than necessary. The following uses essentially the same logic but implements it as a single line while loop.
// Remove trailing spaces from myString
while myString.last == " " { myString = String(myString.dropLast()) }
This can also be written using the .isWhitespace property, as in #demosthese's solution, as follows:
while myString.last?.isWhitespace == true { myString = String(myString.dropLast()) }
This has the benefit (or disadvantage, depending on your point of view) that this removes all types of whitespace, not just spaces but (according to Apple docs) also including newlines, and specifically the following characters:
“\t” (U+0009 CHARACTER TABULATION)
“ “ (U+0020 SPACE)
U+2029 PARAGRAPH SEPARATOR
U+3000 IDEOGRAPHIC SPACE
Note: Even though .isWhitespace is a Boolean it can't be used directly in the while loop as it ends up being optional ? due to the chaining of the optional .last property, which returns nil if the String (or collection) is empty. The == true logic gets around this since nil != true.
I'd love to get some feedback on this, esp. in case anyone sees any issues or drawbacks with this simple single line approach.

Swift 5
extension String {
func trimTrailingWhiteSpace() -> String {
guard self.last == " " else { return self }
var tmp = self
repeat {
tmp = String(tmp.dropLast())
} while tmp.last == " "
return tmp
}
}

Related

Escaping Regex with special characters in Swift

I have a relatively complex regex that I need to run in Swift. Originally was:
"typedef\W+struct\W+{([^}]*)}\W+(\w+);"
You can see the pattern working in JS here.
To make it compile in Swift I escaped the backslashes to:
"typedef\\W+struct\\W+{([^}]*)}\\W+(\\w+);"
On runtime the expression fails to compile with 2048 error. I tried escaping other characters too and tried also escapedPatternForString but without luck. Is there a script to convert JS regexs to Swift? Thanks!
You need to escape both { and } that are outside of a character class:
let rx = "typedef\\W+struct\\W+\\{([^}]*)\\}\\W+(\\w+);"
A quick demo:
let rx = "typedef\\W+struct\\W+\\{([^}]*)\\}\\W+(\\w+);"
let str = "typedef: struct { something } text;"
print(str.range(of: rx, options: .regularExpression) != nil)
// => true
When the { and } are inside a character class they may stay unescaped (as in [^}]).
Using this code (answer by Confused Vorlon), you may get the first match with all capturing groups:
extension NSTextCheckingResult {
func groups(testedString:String) -> [String] {
var groups = [String]()
for i in 0 ..< self.numberOfRanges
{
let group = String(testedString[Range(self.range(at: i), in: testedString)!])
groups.append(group)
}
return groups
}
}
let str = "typedef: struct { something } text;"
let rx = "typedef\\W+struct\\W+\\{([^}]*)\\}\\W+(\\w+);"
let MyRegex = try! NSRegularExpression(pattern: rx)
if let match = MyRegex.firstMatch(in: str, range: NSMakeRange(0, str.count)) {
let groups = match.groups(testedString: str)
print(groups)
}
// => ["typedef: struct { something } text;", " something ", "text"]

Remove the first six characters from a String (Swift)

What's the best way to go about removing the first six characters of a string? Through Stack Overflow, I've found a couple of ways that were supposed to be solutions but I noticed an error with them. For instance,
extension String {
func removing(charactersOf string: String) -> String {
let characterSet = CharacterSet(charactersIn: string)
let components = self.components(separatedBy: characterSet)
return components.joined(separator: "")
}
If I type in a website like https://www.example.com, and store it as a variable named website, then type in the following
website.removing(charactersOf: "https://")
it removes the https:// portion but it also removes all h's, all t's, :'s, etc. from the text.
How can I just delete the first characters?
In Swift 4 it is really simple, just use dropFirst(n: Int)
let myString = "Hello World"
myString.dropFirst(6)
//World
In your case: website.dropFirst(6)
Why not :
let stripped = String(website.characters.dropFirst(6))
Seems more concise and straightforward to me.
(it won't work with multi-char emojis either mind you)
[EDIT] Swift 4 made this even shorter:
let stripped = String(website.dropFirst(6))
length is the number of characters you want to remove (6 in your case)
extension String {
func toLengthOf(length:Int) -> String {
if length <= 0 {
return self
} else if let to = self.index(self.startIndex, offsetBy: length, limitedBy: self.endIndex) {
return self.substring(from: to)
} else {
return ""
}
}
}
It will remove first 6 characters from a string
var str = "Hello-World"
let range1 = str.characters.index(str.startIndex, offsetBy: 6)..<str.endIndex
str = str[range1]
print("the end time is : \(str)")

swift: how can I delete a specific character?

a string such as ! !! yuahl! ! , I want delete ! and , when I code like this
for index in InputName.characters.indices {
if String(InputName[index]) == "" || InputName.substringToIndex(index) == "!" {
InputName.removeAtIndex(index)
}
}
have this error " fatal error: subscript: subRange extends past String end ", how should I do? THX :D
Swift 5+
let myString = "aaaaaaaabbbb"
let replaced = myString.replacingOccurrences(of: "bbbb", with: "") // "aaaaaaaa"
If you need to remove characters only on both ends, you can use stringByTrimmingCharactersInSet(_:)
let delCharSet = NSCharacterSet(charactersInString: "! ")
let s1 = "! aString! !"
let s1Del = s1.stringByTrimmingCharactersInSet(delCharSet)
print(s1Del) //->aString
let s2 = "! anotherString !! aString! !"
let s2Del = s2.stringByTrimmingCharactersInSet(delCharSet)
print(s2Del) //->anotherString !! aString
If you need to remove characters also in the middle, "reconstruct from the filtered output" would be a little bit more efficient than repeating single character removal.
var tempUSView = String.UnicodeScalarView()
tempUSView.appendContentsOf(s2.unicodeScalars.lazy.filter{!delCharSet.longCharacterIsMember($0.value)})
let s2DelAll = String(tempUSView)
print(s2DelAll) //->anotherStringaString
If you don't mind generating many intermediate Strings and Arrays, this single liner can generate the expected output:
let s2DelAll2 = s2.componentsSeparatedByCharactersInSet(delCharSet).joinWithSeparator("")
print(s2DelAll2) //->anotherStringaString
I find that the filter method is a good way to go for this sort of thing:
let unfiltered = "! !! yuahl! !"
// Array of Characters to remove
let removal: [Character] = ["!"," "]
// turn the string into an Array
let unfilteredCharacters = unfiltered.characters
// return an Array without the removal Characters
let filteredCharacters = unfilteredCharacters.filter { !removal.contains($0) }
// build a String with the filtered Array
let filtered = String(filteredCharacters)
print(filtered) // => "yeah"
// combined to a single line
print(String(unfiltered.characters.filter { !removal.contains($0) })) // => "yuahl"
Swift 3
In Swift 3, the syntax is a bit nicer. As a result of the Great Swiftification of the old APIs, the factory method is now called trimmingCharacters(in:). Also, you can construct the CharacterSet as a Set of single-character Strings:
let string = "! !! yuahl! !"
string.trimmingCharacters(in: [" ", "!"]) // "yuahl"
If you have characters in the middle of the string you would like to remove as well, you can use components(separatedBy:).joined():
let string = "! !! yu !ahl! !"
string.components(separatedBy: ["!", " "]).joined() // "yuahl"
H/T #OOPer for the Swift 2 version
func trimLast(character chars: Set<Character>) -> String {
let str: String = String(self.reversed())
guard let index = str.index(where: {!chars.contains($0)}) else {
return self
}
return String((str[index..<str.endIndex]).reversed())
}
Note:
By adding this function in String extension, you can delete the specific character of string at last.
for index in InputName.characters.indices.reversed() {
if String(InputName[index]) == "" || InputName.substringToIndex(index) == "!" {
InputName.removeAtIndex(index)
}
}
Also you can add such very helpful extension :
import Foundation
extension String{
func exclude(find:String) -> String {
return stringByReplacingOccurrencesOfString(find, withString: "", options: .CaseInsensitiveSearch, range: nil)
}
func replaceAll(find:String, with:String) -> String {
return stringByReplacingOccurrencesOfString(find, withString: with, options: .CaseInsensitiveSearch, range: nil)
}
}
you can use this:
for example if you want to remove "%" the percent from 10%
if let i = text.firstIndex(of: "%") {
text.remove(at: i) //10
}

Remove all non-numeric characters from a string in swift

I have the need to parse some unknown data which should just be a numeric value, but may contain whitespace or other non-alphanumeric characters.
Is there a new way of doing this in Swift? All I can find online seems to be the old C way of doing things.
I am looking at stringByTrimmingCharactersInSet - as I am sure my inputs will only have whitespace/special characters at the start or end of the string. Are there any built in character sets I can use for this? Or do I need to create my own?
I was hoping there would be something like stringFromCharactersInSet() which would allow me to specify only valid characters to keep
I was hoping there would be something like stringFromCharactersInSet() which would allow me to specify only valid characters to keep.
You can either use trimmingCharacters with the inverted character set to remove characters from the start or the end of the string. In Swift 3 and later:
let result = string.trimmingCharacters(in: CharacterSet(charactersIn: "0123456789.").inverted)
Or, if you want to remove non-numeric characters anywhere in the string (not just the start or end), you can filter the characters, e.g. in Swift 4.2.1:
let result = string.filter("0123456789.".contains)
Or, if you want to remove characters from a CharacterSet from anywhere in the string, use:
let result = String(string.unicodeScalars.filter(CharacterSet.whitespaces.inverted.contains))
Or, if you want to only match valid strings of a certain format (e.g. ####.##), you could use regular expression. For example:
if let range = string.range(of: #"\d+(\.\d*)?"#, options: .regularExpression) {
let result = string[range] // or `String(string[range])` if you need `String`
}
The behavior of these different approaches differ slightly so it just depends on precisely what you're trying to do. Include or exclude the decimal point if you want decimal numbers, or just integers. There are lots of ways to accomplish this.
For older, Swift 2 syntax, see previous revision of this answer.
let result = string.stringByReplacingOccurrencesOfString("[^0-9]", withString: "", options: NSStringCompareOptions.RegularExpressionSearch, range:nil).stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceCharacterSet())
Swift 3
let result = string.replacingOccurrences( of:"[^0-9]", with: "", options: .regularExpression)
You can upvote this answer.
I prefer this solution, because I like extensions, and it seems a bit cleaner to me. Solution reproduced here:
extension String {
var digits: String {
return components(separatedBy: CharacterSet.decimalDigits.inverted)
.joined()
}
}
You can filter the UnicodeScalarView of the string using the pattern matching operator for ranges, pass a UnicodeScalar ClosedRange from 0 to 9 and initialise a new String with the resulting UnicodeScalarView:
extension String {
private static var digits = UnicodeScalar("0")..."9"
var digits: String {
return String(unicodeScalars.filter(String.digits.contains))
}
}
"abc12345".digits // "12345"
edit/update:
Swift 4.2
extension RangeReplaceableCollection where Self: StringProtocol {
var digits: Self {
return filter(("0"..."9").contains)
}
}
or as a mutating method
extension RangeReplaceableCollection where Self: StringProtocol {
mutating func removeAllNonNumeric() {
removeAll { !("0"..."9" ~= $0) }
}
}
Swift 5.2 • Xcode 11.4 or later
In Swift5 we can use a new Character property called isWholeNumber:
extension RangeReplaceableCollection where Self: StringProtocol {
var digits: Self { filter(\.isWholeNumber) }
}
extension RangeReplaceableCollection where Self: StringProtocol {
mutating func removeAllNonNumeric() {
removeAll { !$0.isWholeNumber }
}
}
To allow a period as well we can extend Character and create a computed property:
extension Character {
var isDecimalOrPeriod: Bool { "0"..."9" ~= self || self == "." }
}
extension RangeReplaceableCollection where Self: StringProtocol {
var digitsAndPeriods: Self { filter(\.isDecimalOrPeriod) }
}
Playground testing:
"abc12345".digits // "12345"
var str = "123abc0"
str.removeAllNonNumeric()
print(str) //"1230"
"Testing0123456789.".digitsAndPeriods // "0123456789."
Swift 4
I found a decent way to get only alpha numeric characters set from a string.
For instance:-
func getAlphaNumericValue() {
var yourString = "123456789!##$%^&*()AnyThingYouWant"
let unsafeChars = CharacterSet.alphanumerics.inverted // Remove the .inverted to get the opposite result.
let cleanChars = yourString.components(separatedBy: unsafeChars).joined(separator: "")
print(cleanChars) // 123456789AnyThingYouWant
}
A solution using the filter function and rangeOfCharacterFromSet
let string = "sld [f]34é7*˜µ"
let alphaNumericCharacterSet = NSCharacterSet.alphanumericCharacterSet()
let filteredCharacters = string.characters.filter {
return String($0).rangeOfCharacterFromSet(alphaNumericCharacterSet) != nil
}
let filteredString = String(filteredCharacters) // -> sldf34é7µ
To filter for only numeric characters use
let string = "sld [f]34é7*˜µ"
let numericSet = "0123456789"
let filteredCharacters = string.characters.filter {
return numericSet.containsString(String($0))
}
let filteredString = String(filteredCharacters) // -> 347
or
let numericSet : [Character] = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
let filteredCharacters = string.characters.filter {
return numericSet.contains($0)
}
let filteredString = String(filteredCharacters) // -> 347
Swift 4
But without extensions or componentsSeparatedByCharactersInSet which doesn't read as well.
let allowedCharSet = NSCharacterSet.letters.union(.whitespaces)
let filteredText = String(sourceText.unicodeScalars.filter(allowedCharSet.contains))
let string = "+1*(234) fds567#-8/90-"
let onlyNumbers = string.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()
print(onlyNumbers) // "1234567890"
or
extension String {
func removeNonNumeric() -> String {
return self.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()
}
}
let onlyNumbers = "+1*(234) fds567#-8/90-".removeNonNumeric()
print(onlyNumbers)// "1234567890"
Swift 3, filters all except numbers
let myString = "dasdf3453453fsdf23455sf.2234"
let result = String(myString.characters.filter { String($0).rangeOfCharacter(from: CharacterSet(charactersIn: "0123456789")) != nil })
print(result)
Swift 4.2
let numericString = string.filter { (char) -> Bool in
return char.isNumber
}
You can do something like this...
let string = "[,myString1. \"" // string : [,myString1. "
let characterSet = NSCharacterSet(charactersInString: "[,. \"")
let finalString = (string.componentsSeparatedByCharactersInSet(characterSet) as NSArray).componentsJoinedByString("")
print(finalString)
//finalString will be "myString1"
The issue with Rob's first solution is stringByTrimmingCharactersInSet only filters the ends of the string rather than throughout, as stated in Apple's documentation:
Returns a new string made by removing from both ends of the receiver characters contained in a given character set.
Instead use componentsSeparatedByCharactersInSet to first isolate all non-occurrences of the character set into arrays and subsequently join them with an empty string separator:
"$$1234%^56()78*9££".componentsSeparatedByCharactersInSet(NSCharacterSet(charactersInString: "0123456789").invertedSet)).joinWithSeparator("")
Which returns 123456789
Swift 3
extension String {
var keepNumericsOnly: String {
return self.components(separatedBy: CharacterSet(charactersIn: "0123456789").inverted).joined(separator: "")
}
}
Swift 4.0 version
extension String {
var numbers: String {
return String(describing: filter { String($0).rangeOfCharacter(from: CharacterSet(charactersIn: "0123456789")) != nil })
}
}
Swift 4
String.swift
import Foundation
extension String {
func removeCharacters(from forbiddenChars: CharacterSet) -> String {
let passed = self.unicodeScalars.filter { !forbiddenChars.contains($0) }
return String(String.UnicodeScalarView(passed))
}
func removeCharacters(from: String) -> String {
return removeCharacters(from: CharacterSet(charactersIn: from))
}
}
ViewController.swift
let character = "1Vi234s56a78l9"
let alphaNumericSet = character.removeCharacters(from: CharacterSet.decimalDigits.inverted)
print(alphaNumericSet) // will print: 123456789
let alphaNumericCharacterSet = character.removeCharacters(from: "0123456789")
print("no digits",alphaNumericCharacterSet) // will print: Vishal
Swift 4.2
let digitChars = yourString.components(separatedBy:
CharacterSet.decimalDigits.inverted).joined(separator: "")
Swift 3 Version
extension String
{
func trimmingCharactersNot(in charSet: CharacterSet) -> String
{
var s:String = ""
for unicodeScalar in self.unicodeScalars
{
if charSet.contains(unicodeScalar)
{
s.append(String(unicodeScalar))
}
}
return s
}
}

How to capitalize the first character of sentence using Swift

I have a String description that holds my sentence and want to capitalize only the first letter. I tried different things but most of them give me exceptions and errors. I'm using Xcode 6.
Here is what I tried so far:
let cap = [description.substringToIndex(advance(0,1))] as String
description = cap.uppercaseString + description.substringFromIndex(1)
It gives me:
Type 'String.Index' does not conform to protocol 'IntegerLiteralConvertible'
I tried:
func capitalizedStringWithLocale(locale:0) -> String
But I haven't figured out how to make it work.
In Swift 2, you can do
String(text.characters.first!).capitalizedString + String(text.characters.dropFirst())
Another possibility in Swift 3:
extension String {
func capitalizeFirst() -> String {
let firstIndex = self.index(startIndex, offsetBy: 1)
return self.substring(to: firstIndex).capitalized + self.substring(from: firstIndex).lowercased()
}
}
For Swift 4:
Warnings from above Swift 3 code:
'substring(to:)' is deprecated: Please use String slicing subscript
with a 'partial range upto' operator.
'substring(from:)' is deprecated: Please use String slicing subscript with a 'partial range from' operator.
Swift 4 solution:
extension String {
var capitalizedFirst: String {
guard !isEmpty else {
return self
}
let capitalizedFirstLetter = charAt(i: 0).uppercased()
let secondIndex = index(after: startIndex)
let remainingString = self[secondIndex..<endIndex]
let capitalizedString = "\(capitalizedFirstLetter)\(remainingString)"
return capitalizedString
}
}
Swift 5.0
Answer 1:
extension String {
func capitalizingFirstLetter() -> String {
return prefix(1).capitalized + dropFirst()
}
mutating func capitalizeFirstLetter() {
self = self.capitalizingFirstLetter()
}
}
Answer 2:
extension String {
func capitalizeFirstLetter() -> String {
return self.prefix(1).capitalized + dropFirst()
}
}
Answer 3:
extension String {
var capitalizeFirstLetter:String {
return self.prefix(1).capitalized + dropFirst()
}
}
import Foundation
// A lowercase string
let description = "the quick brown fox jumps over the lazy dog."
// The start index is the first letter
let first = description.startIndex
// The rest of the string goes from the position after the first letter
// to the end.
let rest = advance(first,1)..<description.endIndex
// Glue these two ranges together, with the first uppercased, and you'll
// get the result you want. Note that I'm using description[first...first]
// to get the first letter because I want a String, not a Character, which
// is what you'd get with description[first].
let capitalised = description[first...first].uppercaseString + description[rest]
// Result: "The quick brown fox jumps over the lazy dog."
You may want to make sure there's at least one character in your sentence before you start, as otherwise you'll get a runtime error trying to advance the index beyond the end of the string.
Here is how to do it in Swift 4; just in case if it helps anybody:
extension String {
func captalizeFirstCharacter() -> String {
var result = self
let substr1 = String(self[startIndex]).uppercased()
result.replaceSubrange(...startIndex, with: substr1)
return result
}
}
It won't mutate the original String.
extension String {
var capitalizedFirstLetter:String {
let string = self
return string.replacingCharacters(in: startIndex...startIndex, with: String(self[startIndex]).capitalized)
}
}
Answer:
let newSentence = sentence.capitalizedFirstLetter
For one or each word in string, you can use String's .capitalized property.
print("foo".capitalized) //prints: Foo
print("foo foo foo".capitalized) //prints: Foo Foo Foo
Swift 4.2 version:
extension String {
var firstCharCapitalized: String {
switch count {
case 0:
return self
case 1:
return uppercased()
default:
return self[startIndex].uppercased() + self[index(after: startIndex)...]
}
}
}
Simplest soulution for Swift 4.0.
Add as a computed property extension:
extension String {
var firstCapitalized: String {
var components = self.components(separatedBy: " ")
guard let first = components.first else {
return self
}
components[0] = first.capitalized
return components.joined(separator: " ")
}
}
Usage:
"hello world".firstCapitalized