removing white spaces at beginning and end of string - ios

I've got a problem with removing whitespaces at the beginning and end of string. For e.g. I've got a string like:
\r\n\t- Someone will come here?\n- I don't know for sure...\r\n\r\n
And I need to remove whitespaces only at the end and beginning (string should be look like:
- Someone will come here?\n- I don't know for sure...
Also there could be a lot of variants of string end: "\r\n\r\n", "\r\n", "\n\r\n" and so on...
Thanks.

Your string contains not only whitespace but also new line characters.
Use stringByTrimmingCharactersInSet with whitespaceAndNewlineCharacterSet.
let string = "\r\n\t- Someone will come here?\n- I don't know for sure...\r\n\r\n"
let trimmedString = string.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet())
In Swift 3 it's more cleaned up:
let trimmedString = string.trimmingCharacters(in: .whitespacesAndNewlines)

String trimming first and last whitespaces and newlines in Swift 4+
" 2 space ".trimmingCharacters(in: .whitespacesAndNewlines)
result:
"2 space"

//here i have used regular expression and replaced white spaces at the start and end
let stringPassing : NSString? = "hdfjkhsdj hfjksdhf sdf "
do {
print("old->\(stringPassing)")
let pattern : String = "(^\\s+)|(\\s+)$"
let regex = try NSRegularExpression(pattern: pattern , options: [])
let newMatched = regex.matchesInString(stringPassing! as String, options: [], range: NSMakeRange(0,stringPassing!.length))
if(newMatched.count > 0){
let modifiedString = regex.stringByReplacingMatchesInString(stringPassing! as String, options: [] , range: NSMakeRange(0,stringPassing!.length), withTemplate: "")
print("new->\(modifiedString)")
}
} catch let error as NSError {
print(error.localizedDescription)
}

You can use this extension and just call "yourString".trim()
extension String
{
func trim() -> String
{
return self.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet())
}
}

If you want to remove Whitespace from the start of the string Try this Code.
func removeSpace()->String{
let txt = myString
var count:Int! = 0
for i in txt{
if i == " "{
count += 1
}else{
if count != 0{
return String(txt.dropFirst(count))
}else{
return txt
}
}
}
return ""
}
`

Remove all whiteSpaces from the start of string
func removeWhiteSpaces(str:String) -> String{
var newStr = str
for i in 0..<str.count{
let index = str.index(str.startIndex, offsetBy: i)
print(str[index])
if str[index] != " "{
return newStr
}
else{
newStr.remove(at: newStr.startIndex)
}
}
return newStr
}

In this code to restrict Textfield beginning white space
func textField(_ textField: UITextField, shouldChangeCharactersIn range: NSRange, replacementString string: String) -> Bool {
if (textField.text?.count)! == 0 && string == " " {
return false
}
else{
return true
}
}

In Swift 4
Use it on any String type variable.
extension String {
func trimWhiteSpaces() -> String {
let whiteSpaceSet = NSCharacterSet.whitespaces
return self.trimmingCharacters(in: whiteSpaceSet)
}
}
And Call it like this
yourString.trimWhiteSpaces()

You could first remove from the beginning and then from the end like so:
while true {
if sentence.characters.first == "\r\n" || sentence.characters.first == "\n" {
sentence.removeAtIndex(sentence.startIndex)
} else {
break
}
}
while true {
if sentence.characters.last == "\r\n" || sentence.characters.last == "\n" {
sentence.removeAtIndex(sentence.endIndex.predecessor())
} else {
break
}
}

let string = " exam ple "
let trimmed = string.replacingOccurrences(of: "(^\s+)|(\s+)$", with: "", options: .regularExpression)
print(">" + trimmed3 + "<")
// prints >exam ple<

Related

Get last sentence from String after ". "(Dot And Space)

I am trying to get the last sentence from String after ". "(Dot And Space), Suppose I have a very long string and in that string, there are multiple ". "(Dot And Space) but I want to fetch the String after the very last ". "(Dot And Space) I have tried some of the solutions but I am getting below error
Example String: "Hello. playground!. Hello. World!!! How Are you All?"
Expected Output Needed: "World!!! How Are you All?"
Below is the code which I have tried so far
let fullstring = "Hello. playground!. Hello. World!!! How Are you All?"
let outputString = fullstring.substringAfterLastOccurenceOf(". ") // Here I am getting this error Cannot convert value of type 'String' to expected argument type 'Character'
extension String {
var nsRange: NSRange {
return Foundation.NSRange(startIndex ..< endIndex, in: self)
}
subscript(nsRange: NSRange) -> Substring? {
return Range(nsRange, in: self)
.flatMap { self[$0] }
}
func substringAfterLastOccurenceOf(_ char: Character) -> String {
let regex = try! NSRegularExpression(pattern: "\(char)\\s*(\\S[^\(char)]*)$")
if let match = regex.firstMatch(in: self, range: self.nsRange), let result = self[match.range(at: 1)] {
return String(result)
}
return ""
}
}
Can anyone help me with this or is there any other way to achieve this?
Thanks in Advance!!
Using components(separatedBy:) instead
let fullstring = "Hello. playground!. Hello. World!!! How Are you All?"
let sliptArr = fullstring.components(separatedBy: ". ")
print(sliptArr[sliptArr.count - 1]) // World!!! How Are you All?
if you are using swift 5.7+ and iOS 16.0+ u can use ```regexBuilder'''.
let text = "Hello. playground!. Hello. World!!! How Are you All?"
let regex = Regex {
Optionally{
ZeroOrMore(.any)
". "
}
Capture {
ZeroOrMore(.any)
}
}
if let lastSentence = try? regex.wholeMatch(in: text) {
print("lastSentence: \(lastSentence.output.1)")
}

Textfield Validation using CharacterSet

I have written below code to validate text input in textfield.
else if (textField == txtField_Password)
{
let charSet = CharacterSet(charactersIn: "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789#$&*!")
let charLength = (txtField_Password.text!.count) + (string.count) - range.length
for i in 0..<string.count
{
let c = (string as NSString).character(at: i)
if (!((charSet as NSCharacterSet).characterIsMember(c)))
{
return false
}
}
return (charLength > 20) ? false : true
}
Can anyone help me to convert character(at:) and characterIsMember() part to its swift equivalent in the above code.
You can simplify the logic just by checking the range of the inverted character set. If the string contains only allowed characters the function returns nil.
else if textField == txtField_Password {
let charLength = txtField_Password.text!.utf8.count + string.utf8.count - range.length
let charSet = CharacterSet(charactersIn: "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789#$&*!")
return string.rangeOfCharacter(from: charSet.inverted) == nil && charLength < 21
}
Note that there is a simpler way to implement what you want using a regular expression:
let currentText = (textField.text ?? "") as NSString
let newText = currentText.replacingCharacters(in: range, with: string)
let pattern = "^[a-zA-Z0-9#$&*!]{0,20}$"
return newText.range(of: pattern, options: .regularExpression) != nil
You could work with something along these lines. I appreciate this is a bit rough and ready but should work:
charSet = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789#$&*!"
if txtField_Password.text!.count <= 20 {
for i in 0..<str.count
{
let c = Array(str)[i]
let cString = String(c)
if charSet.contains(cString) {
return false
}
}
} else {
return false
}
Use rangeOfCharacter:
func textField(_ textField: UITextField, shouldChangeCharactersIn range: NSRange, replacementString string: String) -> Bool {
let specialCharacters = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789#$&*!"
let characterSet = CharacterSet(charactersIn: specialCharacters)
guard let lengh = textfield.text else {return}
if lengh.count >= 20 {
// text exceeded 20 characters. Do something
}
if (string.rangeOfCharacter(from: characterSet) != nil) {
print("matched")
return true
} else {
print("not matched")
}
return true
}

Replace space with underscore automatically using Swift [duplicate]

I am looking for a way to replace characters in a Swift String.
Example: "This is my string"
I would like to replace " " with "+" to get "This+is+my+string".
How can I achieve this?
This answer has been updated for Swift 4 & 5. If you're still using Swift 1, 2 or 3 see the revision history.
You have a couple of options. You can do as #jaumard suggested and use replacingOccurrences()
let aString = "This is my string"
let newString = aString.replacingOccurrences(of: " ", with: "+", options: .literal, range: nil)
And as noted by #cprcrack below, the options and range parameters are optional, so if you don't want to specify string comparison options or a range to do the replacement within, you only need the following.
let aString = "This is my string"
let newString = aString.replacingOccurrences(of: " ", with: "+")
Or, if the data is in a specific format like this, where you're just replacing separation characters, you can use components() to break the string into and array, and then you can use the join() function to put them back to together with a specified separator.
let toArray = aString.components(separatedBy: " ")
let backToString = toArray.joined(separator: "+")
Or if you're looking for a more Swifty solution that doesn't utilize API from NSString, you could use this.
let aString = "Some search text"
let replaced = String(aString.map {
$0 == " " ? "+" : $0
})
You can use this:
let s = "This is my string"
let modified = s.replace(" ", withString:"+")
If you add this extension method anywhere in your code:
extension String
{
func replace(target: String, withString: String) -> String
{
return self.stringByReplacingOccurrencesOfString(target, withString: withString, options: NSStringCompareOptions.LiteralSearch, range: nil)
}
}
Swift 3:
extension String
{
func replace(target: String, withString: String) -> String
{
return self.replacingOccurrences(of: target, with: withString, options: NSString.CompareOptions.literal, range: nil)
}
}
Swift 3, Swift 4, Swift 5 Solution
let exampleString = "Example string"
//Solution suggested above in Swift 3.0
let stringToArray = exampleString.components(separatedBy: " ")
let stringFromArray = stringToArray.joined(separator: "+")
//Swiftiest solution
let swiftyString = exampleString.replacingOccurrences(of: " ", with: "+")
Did you test this :
var test = "This is my string"
let replaced = test.stringByReplacingOccurrencesOfString(" ", withString: "+", options: nil, range: nil)
var str = "This is my string"
print(str.replacingOccurrences(of: " ", with: "+"))
Output is
This+is+my+string
Swift 5.5
I am using this extension:
extension String {
func replaceCharacters(characters: String, toSeparator: String) -> String {
let characterSet = CharacterSet(charactersIn: characters)
let components = components(separatedBy: characterSet)
let result = components.joined(separator: toSeparator)
return result
}
func wipeCharacters(characters: String) -> String {
return self.replaceCharacters(characters: characters, toSeparator: "")
}
}
Usage:
"<34353 43434>".replaceCharacters(characters: "< >", toSeparator:"+") // +34353+43434+
"<34353 43434>".wipeCharacters(characters: "< >") // 3435343434
Swift 4:
let abc = "Hello world"
let result = abc.replacingOccurrences(of: " ", with: "_",
options: NSString.CompareOptions.literal, range:nil)
print(result :\(result))
Output:
result : Hello_world
A Swift 3 solution along the lines of Sunkas's:
extension String {
mutating func replace(_ originalString:String, with newString:String) {
self = self.replacingOccurrences(of: originalString, with: newString)
}
}
Use:
var string = "foo!"
string.replace("!", with: "?")
print(string)
Output:
foo?
A category that modifies an existing mutable String:
extension String
{
mutating func replace(originalString:String, withString newString:String)
{
let replacedString = self.stringByReplacingOccurrencesOfString(originalString, withString: newString, options: nil, range: nil)
self = replacedString
}
}
Use:
name.replace(" ", withString: "+")
Swift 3 solution based on Ramis' answer:
extension String {
func withReplacedCharacters(_ characters: String, by separator: String) -> String {
let characterSet = CharacterSet(charactersIn: characters)
return components(separatedBy: characterSet).joined(separator: separator)
}
}
Tried to come up with an appropriate function name according to Swift 3 naming convention.
Less happened to me, I just want to change (a word or character) in the String
So I've use the Dictionary
extension String{
func replace(_ dictionary: [String: String]) -> String{
var result = String()
var i = -1
for (of , with): (String, String)in dictionary{
i += 1
if i<1{
result = self.replacingOccurrences(of: of, with: with)
}else{
result = result.replacingOccurrences(of: of, with: with)
}
}
return result
}
}
usage
let mobile = "+1 (800) 444-9999"
let dictionary = ["+": "00", " ": "", "(": "", ")": "", "-": ""]
let mobileResult = mobile.replace(dictionary)
print(mobileResult) // 001800444999
Xcode 11 • Swift 5.1
The mutating method of StringProtocol replacingOccurrences can be implemented as follow:
extension RangeReplaceableCollection where Self: StringProtocol {
mutating func replaceOccurrences<Target: StringProtocol, Replacement: StringProtocol>(of target: Target, with replacement: Replacement, options: String.CompareOptions = [], range searchRange: Range<String.Index>? = nil) {
self = .init(replacingOccurrences(of: target, with: replacement, options: options, range: searchRange))
}
}
var name = "This is my string"
name.replaceOccurrences(of: " ", with: "+")
print(name) // "This+is+my+string\n"
var str = "This is my string"
str = str.replacingOccurrences(of: " ", with: "+")
print(str)
This is easy in swift 4.2. just use replacingOccurrences(of: " ", with: "_") for replace
var myStr = "This is my string"
let replaced = myStr.replacingOccurrences(of: " ", with: "_")
print(replaced)
Since Swift 2, String does no longer conform to SequenceType. In other words, you can not iterate through a string with a for...in loop.
The simple and easy way is to convert String to Array to get the benefit of the index just like that:
let input = Array(str)
I remember when I tried to index into String without using any conversion. I was really frustrated that I couldn’t come up with or reach a desired result, and was about to give up.
But I ended up creating my own workaround solution, and here is the full code of the extension:
extension String {
subscript (_ index: Int) -> String {
get {
String(self[self.index(startIndex, offsetBy: index)])
}
set {
remove(at: self.index(self.startIndex, offsetBy: index))
insert(Character(newValue), at: self.index(self.startIndex, offsetBy: index))
}
}
}
Now that you can read and replace a single character from string using its index just like you originally wanted to:
var str = "cat"
for i in 0..<str.count {
if str[i] == "c" {
str[i] = "h"
}
}
print(str)
It’s simple and useful way to use it and get through Swift’s String access model.
Now that you’ll feel it’s smooth sailing next time when you can loop through the string just as it is, not casting it into Array.
Try it out, and see if it can help!
I've implemented this very simple func:
func convap (text : String) -> String {
return text.stringByReplacingOccurrencesOfString("'", withString: "''")
}
So you can write:
let sqlQuery = "INSERT INTO myTable (Field1, Field2) VALUES ('\(convap(value1))','\(convap(value2)')
I think Regex is the most flexible and solid way:
var str = "This is my string"
let regex = try! NSRegularExpression(pattern: " ", options: [])
let output = regex.stringByReplacingMatchesInString(
str,
options: [],
range: NSRange(location: 0, length: str.characters.count),
withTemplate: "+"
)
// output: "This+is+my+string"
Swift extension:
extension String {
func stringByReplacing(replaceStrings set: [String], with: String) -> String {
var stringObject = self
for string in set {
stringObject = self.stringByReplacingOccurrencesOfString(string, withString: with)
}
return stringObject
}
}
Go on and use it like let replacedString = yorString.stringByReplacing(replaceStrings: [" ","?","."], with: "+")
The speed of the function is something that i can hardly be proud of, but you can pass an array of String in one pass to make more than one replacement.
Here is the example for Swift 3:
var stringToReplace = "This my string"
if let range = stringToReplace.range(of: "my") {
stringToReplace?.replaceSubrange(range, with: "your")
}
Here's an extension for an in-place occurrences replace method on String, that doesn't no an unnecessary copy and do everything in place:
extension String {
mutating func replaceOccurrences<Target: StringProtocol, Replacement: StringProtocol>(of target: Target, with replacement: Replacement, options: String.CompareOptions = [], locale: Locale? = nil) {
var range: Range<Index>?
repeat {
range = self.range(of: target, options: options, range: range.map { self.index($0.lowerBound, offsetBy: replacement.count)..<self.endIndex }, locale: locale)
if let range = range {
self.replaceSubrange(range, with: replacement)
}
} while range != nil
}
}
(The method signature also mimics the signature of the built-in String.replacingOccurrences() method)
May be used in the following way:
var string = "this is a string"
string.replaceOccurrences(of: " ", with: "_")
print(string) // "this_is_a_string"
If you don't want to use the Objective-C NSString methods, you can just use split and join:
var string = "This is my string"
string = join("+", split(string, isSeparator: { $0 == " " }))
split(string, isSeparator: { $0 == " " }) returns an array of strings (["This", "is", "my", "string"]).
join joins these elements with a +, resulting in the desired output: "This+is+my+string".
you can test this:
let newString = test.stringByReplacingOccurrencesOfString(" ", withString: "+", options: nil, range: nil)
Swift 5.5
But this might work in earlier versions.
I'm frequently replacing because I want to replace "any whitespace or -" with a _ or something like that. This extension on string lets me do that.
extension String {
func removingCharacters(_ characters:CharacterSet) -> Self {
Self(self.unicodeScalars.filter {
!characters.contains($0)
})
}
func removingCharacters(in string:String) -> Self {
Self(self.unicodeScalars.filter {
!CharacterSet(charactersIn:string).contains($0)
})
}
func replacingCharacters(_ characters:CharacterSet, with newChar:Character) -> Self {
String(self.compactMap( {
CharacterSet(charactersIn: "\($0.1)").isSubset(of: characters)
? newChar : $0.1
}))
}
func replacingCharacters(in string:String, with newChar:Character) -> Self {
String(self.compactMap( {
CharacterSet(charactersIn: "\($0)").isSubset(of: CharacterSet(charactersIn:string))
? newChar : $0
}))
}
}
usage:
print("hello \n my name\t is Joe".removingCharacters(.whitespacesAndNewlines))
print("hello \n my name\t is Joe".removingCharacters(in: " \t\n"))
print("ban annan anann ana".replacingCharacters(.whitespacesAndNewlines, with: "_"))
print("ban-annan anann ana".replacingCharacters(in: " -", with: "_"))
Obviously for a single character the .replacingOccurrences(of: " ", with: "+") is better.
I have not done a performance comparison to the
let toArray = aString.components(separatedBy: characterSet)
let backToString = toArray.joined(separator: "+")
style done in Ramis's extension. I'd be interested if someone does.
See also replacing emoji's: https://stackoverflow.com/a/63416058/5946596

Remove all occurrences except the last one

I want to remove all occurrences of char == . in a string except the last one.
E.G:
1.2.3.4
should become:
123.4
Find the position of the last dot.
Remove all dots preceding this position.
Example:
var str = "1.2.3.4"
if let idx = str.range(of: ".", options: .backwards) {
str = str.replacingOccurrences(of: ".", with: "", range: str.startIndex..<idx.lowerBound)
}
print(str) // 123.4
Here is the example code. Split by your character, add that character to the beginning of the last element, and then join them back
let str = "1.2.3.4"
var array = str.components(separatedBy: ".")
if array.count >= 1 {
array[array.count - 1] = "." + array[array.count - 1]
}
print(array.joined(separator: ""))
You could make this easilt reusable as an extension on String.
public extension String {
func stringByReplacingAllButLastOccurrenceOfString(target: String, withString replaceString: String) -> String {
if let idx = self.rangeOfString(target, options: .BackwardsSearch) {
return self.stringByReplacingOccurrencesOfString(target, withString: replaceString, range: self.startIndex..<idx.first!)
}
return self
}
}
Running through a few of the scenarios, you get:
strResult = "1.2.3.4.5".stringByReplacingAllButLastOccurrenceOfString(".", withString: "")
// result "1234.5"
strResult = "1234.5".stringByReplacingAllButLastOccurrenceOfString(".", withString: "")
// result "1234.5"
strResult = "1.23.4.5".stringByReplacingAllButLastOccurrenceOfString(".", withString: "")
// result "1234.5"
strResult = "1234.5".stringByReplacingAllButLastOccurrenceOfString(".", withString: "")
// result "1234.5"
strResult = "12345".stringByReplacingAllButLastOccurrenceOfString(".", withString: "")
// result "12345"

Separating CamelCase string into space-separated words in Swift

I would like to separate a CamelCase string into space-separated words in a new string. Here is what I have so far:
var camelCaps: String {
guard self.count > 0 else { return self }
var newString: String = ""
let uppercase = CharacterSet.uppercaseLetters
let first = self.unicodeScalars.first!
newString.append(Character(first))
for scalar in self.unicodeScalars.dropFirst() {
if uppercase.contains(scalar) {
newString.append(" ")
}
let character = Character(scalar)
newString.append(character)
}
return newString
}
let aCamelCaps = "aCamelCaps"
let camelCapped = aCamelCaps.camelCaps // Produce: "a Camel Caps"
let anotherCamelCaps = "ÄnotherCamelCaps"
let anotherCamelCapped = anotherCamelCaps.camelCaps // "Änother Camel Caps"
I'm inclined to suspect that this may not be the most efficient way to convert to space-separated words, if I call it in a tight loop, or 1000's of times. Are there more efficient ways to do this in Swift?
[Edit 1:] The solution I require should remain general for Unicode scalars, not specific to Roman ASCII "A..Z".
[Edit 2:] The solution should also skip the first letter, i.e. not prepend a space before the first letter.
[Edit 3:] Updated for Swift 4 syntax, and added caching of uppercaseLetters, which improves performance in very long strings and tight loops.
extension String {
func camelCaseToWords() -> String {
return unicodeScalars.dropFirst().reduce(String(prefix(1))) {
return CharacterSet.uppercaseLetters.contains($1)
? $0 + " " + String($1)
: $0 + String($1)
}
}
}
print("ÄnotherCamelCaps".camelCaseToWords()) // Änother Camel Caps
May be helpful for someone :)
One Line Solution
I concur with #aircraft, regular expressions can solve this problem in one LOC!
// Swift 5 (and probably 4?)
extension String {
func titleCase() -> String {
return self
.replacingOccurrences(of: "([A-Z])",
with: " $1",
options: .regularExpression,
range: range(of: self))
.trimmingCharacters(in: .whitespacesAndNewlines)
.capitalized // If input is in llamaCase
}
}
Props to this JS answer.
P.S. I have a gist for snake_case → CamelCase here.
P.P.S. I updated this for New Swift (currently 5.1), then saw #busta's answer, and swapped out my startIndex..<endIndex for his range(of: self). Credit where it's due y'all!
a better full swifty solution... based on AmitaiB answer
extension String {
func titlecased() -> String {
return self.replacingOccurrences(of: "([A-Z])", with: " $1", options: .regularExpression, range: self.range(of: self))
.trimmingCharacters(in: .whitespacesAndNewlines)
.capitalized
}
}
I might be late but I want to share a little improvement to Augustine P A answer or Leo Dabus comment.
Basically, that code won't work properly if we are using upper camel case notation (like "DuckDuckGo") because it will add a space at the beginning of the string.
To address this issue, this is a slightly modified version of the code, using Swift 3.x, and it's compatible with both upper and lower came case:
extension String {
func camelCaseToWords() -> String {
return unicodeScalars.reduce("") {
if CharacterSet.uppercaseLetters.contains($1) {
if $0.count > 0 {
return ($0 + " " + String($1))
}
}
return $0 + String($1)
}
}
}
As far as I tested on my old MacBook, your code seems to be efficient enough for short strings:
import Foundation
extension String {
var camelCaps: String {
var newString: String = ""
let upperCase = CharacterSet.uppercaseLetters
for scalar in self.unicodeScalars {
if upperCase.contains(scalar) {
newString.append(" ")
}
let character = Character(scalar)
newString.append(character)
}
return newString
}
var camelCaps2: String {
var newString: String = ""
let upperCase = CharacterSet.uppercaseLetters
var range = self.startIndex..<self.endIndex
while let foundRange = self.rangeOfCharacter(from: upperCase,range: range) {
newString += self.substring(with: range.lowerBound..<foundRange.lowerBound)
newString += " "
newString += self.substring(with: foundRange)
range = foundRange.upperBound..<self.endIndex
}
newString += self.substring(with: range)
return newString
}
var camelCaps3: String {
struct My {
static let regex = try! NSRegularExpression(pattern: "[A-Z]")
}
return My.regex.stringByReplacingMatches(in: self, range: NSRange(0..<self.utf16.count), withTemplate: " $0")
}
}
let aCamelCaps = "aCamelCaps"
assert(aCamelCaps.camelCaps == aCamelCaps.camelCaps2)
assert(aCamelCaps.camelCaps == aCamelCaps.camelCaps3)
let t0 = Date().timeIntervalSinceReferenceDate
for _ in 0..<1_000_000 {
let aCamelCaps = "aCamelCaps"
let camelCapped = aCamelCaps.camelCaps
}
let t1 = Date().timeIntervalSinceReferenceDate
print(t1-t0) //->4.78703999519348
for _ in 0..<1_000_000 {
let aCamelCaps = "aCamelCaps"
let camelCapped = aCamelCaps.camelCaps2
}
let t2 = Date().timeIntervalSinceReferenceDate
print(t2-t1) //->10.5831440091133
for _ in 0..<1_000_000 {
let aCamelCaps = "aCamelCaps"
let camelCapped = aCamelCaps.camelCaps3
}
let t3 = Date().timeIntervalSinceReferenceDate
print(t3-t2) //->14.2085000276566
(Do not try to test the code above in the Playground. The numbers are taken from a single trial executed as a CommandLine app.)
extension String {
func titlecased() -> String {
return self
.replacingOccurrences(of: "([a-z])([A-Z](?=[A-Z])[a-z]*)", with: "$1 $2", options: .regularExpression)
.replacingOccurrences(of: "([A-Z])([A-Z][a-z])", with: "$1 $2", options: .regularExpression)
.replacingOccurrences(of: "([a-z])([A-Z][a-z])", with: "$1 $2", options: .regularExpression)
.replacingOccurrences(of: "([a-z])([A-Z][a-z])", with: "$1 $2", options: .regularExpression)
}
}
In
"ThisStringHasNoSpacesButItDoesHaveCapitals"
"IAmNotAGoat"
"LOLThatsHilarious!"
"ThisIsASMSMessage"
Out
"This String Has No Spaces But It Does Have Capitals"
"I Am Not A Goat"
"LOL Thats Hilarious!"
"This Is ASMS Message" // (Difficult tohandle single letter words when they are next to acronyms.)
enter link description here
I can do this extension in less lines of code (and without a CharacterSet), but yes, you basically have to enumerate each String if you want to insert spaces in front of capital letters.
extension String {
var differentCamelCaps: String {
var newString: String = ""
for eachCharacter in self {
if "A"..."Z" ~= eachCharacter {
newString.append(" ")
}
newString.append(eachCharacter)
}
return newString
}
}
print("ÄnotherCamelCaps".differentCamelCaps) // Änother Camel Caps
Swift 5 solution
extension String {
func camelCaseToWords() -> String {
return unicodeScalars.reduce("") {
if CharacterSet.uppercaseLetters.contains($1) {
if $0.count > 0 {
return ($0 + " " + String($1))
}
}
return $0 + String($1)
}
}
}
If you want to make it more efficient, you can use Regular Expressions.
extension String {
func replace(regex: NSRegularExpression, with replacer: (_ match:String)->String) -> String {
let str = self as NSString
let ret = str.mutableCopy() as! NSMutableString
let matches = regex.matches(in: str as String, options: [], range: NSMakeRange(0, str.length))
for match in matches.reversed() {
let original = str.substring(with: match.range)
let replacement = replacer(original)
ret.replaceCharacters(in: match.range, with: replacement)
}
return ret as String
}
}
let camelCaps = "aCamelCaps" // there are 3 Capital character
let pattern = "[A-Z]"
let regular = try!NSRegularExpression(pattern: pattern)
let camelCapped:String = camelCaps.replace(regex: regular) { " \($0)" }
print("Uppercase characters replaced: \(camelCapped)")
Here's what I came up with using Unicode character classes: (Swift 5)
extension String {
var titleCased: String {
self
.replacingOccurrences(of: "(\\p{UppercaseLetter}\\p{LowercaseLetter}|\\p{UppercaseLetter}+(?=\\p{UppercaseLetter}))",
with: " $1",
options: .regularExpression,
range: range(of: self)
)
.capitalized
}
}
Output:
fillPath ➝ Fill Path
ThisStringHasNoSpaces ➝ This String Has No Spaces
IAmNotAGoat ➝ I Am Not A Goat
LOLThatsHilarious! ➝ Lol Thats Hilarious!
ThisIsASMSMessage ➝ This Is Asms Message
Swift way:
extension String {
var titlecased: String {
map { ($0.isUppercase ? " " : "") + String($0) }
.joined(separator: "")
.trimmingCharacters(in: .whitespaces)
}
}
Swift 5+
Small style improvements on previous answers
import Foundation
extension String {
func camelCaseToWords() -> String {
unicodeScalars.reduce("") {
guard CharacterSet.uppercaseLetters.contains($1),
$0.count > 0
else { return $0 + String($1) }
return ($0 + " " + String($1))
}
}
}
Using guard let statements is usually recommended, as they provide an "early exit" for non matching cases and decrease the overall nesting levels of your code (which usually improves readability quite a lot... and remember, readability counts!)
Solution with REGEX
let camelCase = "SomeATMInTheShop"
let regexPattern = "[A-Z-_&](?=[a-z0-9]+)|[A-Z-_&]+(?![a-z0-9])"
let newValue = camelCase.replacingOccurrences(of: regexPattern, with: " $0", options: .regularExpression, range: nil)
Otuput ==> Some ATM In The Shop

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