Inputting a value with 2 spaces and a string would enable the button " test"
but my code already detects " " whitespaces where the button is disabled but adding additional string on the whitespace enables the button
func validateAccountName(with accountName: String) -> Bool {
let regex = "[^A-Za-zA-O-o-y]"
let accountName = accountName.replacingOccurrences(of: " ", with: "")
return accountName.isEmpty == false && accountName.range(of: regex, options: .regularExpression) == nil
}
func edited() {
// check user input and allow update button to be enabled
guard let inputText = textField.text, let viewModel = viewModel else {
return
}
if !inputText.isEmpty
&& inputText != viewModel.accountName
&& viewModel.validateAccountName(with: inputText) {
self.navigationItem.rightBarButtonItem?.isEnabled = true
} else {
self.navigationItem.rightBarButtonItem?.isEnabled = false
}
}
Well what do you want to achieve?
If the goal is to reject everything that has any whitespaces you can use:
accountName.rangeOfCharacter(from: .whitespaces) == nil
But your regex already checks this as only Characters A-Z, a-z, A-O, - and o-y are allowed.
Maybe this is already what you want?
func validateAccountName(with accountName: String) -> Bool {
let regex = "[^A-Za-z]"
return accountName.range(of: regex, options: .regularExpression) == nil
}
everyone, I need to live update character by character a String/TextField depending on the length of the String.
Eg: If user inserts 43041058, it should change to 4.304.105-8,
else if user inserts one more number (12 total characters) 430410582, it should update to 43.041.058-2.
Real Time Execution eg:
Step | User Input
1 | 4.
2 | 4.3
3 | 4.30
4 | 4.304.
5 | 4.304.1
6 | 4.304.10
7 | 4.304.105
8 | 4.304.105-8
9 | 43.041.058-2
The important thing is that the textField updates itself for each Character pressed, the closet I've got its to make it work only for 12 positions not for 11.
override func viewDidLoad() {
super.viewDidLoad()
rutFormatter()
}
func rutFormatter() {
textRut.addTarget(self, action: #selector(VCLogin.textFieldDidChange(_:)), for: UIControl.Event.editingChanged)
}
#objc func textFieldDidChange(_ textField: UITextField) {
var newRut = String()
for (index, character) in (textField.text?.enumerated())! {
if index % 2 == 0 {
if index != 0 && index % 2 == 0 {
switch index {
case 2:
if character != "." {
newRut.append(".")
}
case 10:
if character != "-" {
newRut.append("-")
}
default:
print("Default")
}
}
if index != 0 && index % 3 == 0 {
switch index {
case 6:
if character != "." {
newRut.append(".")
}
default:
print("Default")
}
}
}
newRut.append(String(character))
}
textField.text = newRut
}
The following code should work.
let mask = "X.XXX.XXX-X"
let mask2 = "XX.XXX.XXX-X"
#objc func textFieldDidChange(_ textField: UITextField) {
let string = textField.text
let number = formattedNumber(number: textField.text!)
textField.text = number
print(string)
}
private func formattedNumber(number: String) -> String {
var cleanPhoneNumber = number.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()
var result = ""
var index = cleanPhoneNumber.startIndex
var fieldMask: String!
if number.count < 12 {
fieldMask = mask
}else {
fieldMask = mask2
}
for ch in fieldMask.characters {
if index == cleanPhoneNumber.endIndex {
break
}
if ch == "X" {
result.append(cleanPhoneNumber[index])
index = cleanPhoneNumber.index(after: index)
} else {
result.append(ch)
}
}
return result
}
In Swift i am found one Unicode so I Have use one check if textfield have this unicode "\u{ef}" condition is false
Here is my code
But is not Working
let getMessage = self.txtChat_View?.text?.trimmingCharacters(in: NSCharacterSet.whitespacesAndNewlines)
if(String(describing: getMessage?.characters) == "\u{ef}")
{
print("gdfsds")
}
print(String(describing: getMessage?.characters)) //ANS Blank
print("-------------------------------")
print("---------->",getMessage ?? String()) //ANS Blank
print(getMessage?.characters.count) //ANS count is 1
if(getMessage ?? String() == "\u{ef}")
{
print("This is unicode")
}
if(getMessage == "")
{
print("This is null")
}
if(getMessage == nil)
{
print("####NIL#####")
}
The original question was "The function beginsWithVowel should take a single String parameter and return a Bool indicating whether the input string begins with a vowel. If the input string begins with a vowel return true, otherwise return false."
func lowercase(a: String) ->String{
return a.lowercaseString
}
func lowercase(a: String) ->String{
return a.lowercaseString
}
func beginsWithVowel(a: String) ->Bool {
if a.characters[a.startIndex] != ("a") && a.characters[a.startIndex] != ("e") && a.characters[a.startIndex] != ("i") && a.characters[a.startIndex] != ("o") && a.characters[a.startIndex] != ("u") {
print("The word must start with a vowel letter.")
return false
}else {
print("Succes!")
return true
}
}
When the a = ""
beginsWithVowel(lowercase(""))
An error occurred.
What should I add to make the function say reminder sentence instead of an error?
I have tried to add those into my code, but the error still occurred(ps: the func lowercase was added after fails)
a.characters[a.startIndex] != ("")
and
if a.characters.count == 0 {
}
You can simply return false if your string is empty otherwise create a string with the vowels and check if it contains the first character of your string:
Swift 3
func beginsWithVowel(a: String) -> Bool {
return a.isEmpty ? false : "aeiouAEIOU".characters.contains(a.characters.first!)
}
Swift 4
func beginsWithVowel(a: String) ->Bool {
return a.isEmpty ? false : "aeiouAEIOU".contains(a.first!)
}
beginsWithVowel(a: "Apple") // true
Note that it will return false for vowels with accent. If you would like to make your method diacritic insensitive you can use string's method func folding(options: String.CompareOptions = default, locale: Locale?) -> String to return the string without accent for comparison:
Swift 3
func beginsWithVowel(a: String) ->Bool {
return a.isEmpty ? false : "aeiouAEIOU".characters.contains(a.folding(options: .diacriticInsensitive, locale: nil).characters.first!)
}
Swift 4
func beginsWithVowel(a: String) ->Bool {
return a.isEmpty ? false : "aeiouAEIOU".contains(a.folding(options: .diacriticInsensitive, locale: nil).first!)
}
beginsWithVowel(a: "águia") // true
I like many of the features in Swift, but using manipulating strings are still a big pain in the ass.
func checkPalindrome(word: String) -> Bool {
print(word)
if word == "" {
return true
} else {
if word.characters.first == word.characters.last {
return checkPalindrome(word.substringWithRange(word.startIndex.successor() ..< word.endIndex.predecessor()))
} else {
return false
}
}
}
This code fails miserably whenever the string's length is an odd number. Of course I could make it so the first line of the block would be if word.characters.count < 2, but is there a way in Swift to get substrings and check easily?
Update
I like many of the suggestions, but I guess the original question could be misleading a little, since it's a question about String more than getting the right results for the function.
For instance, in Python, checkPalindrome(word[1:-1]) would work fine for the recursive definition, whereas Swift code is much less graceful since it needs other bells and whistles.
return word == String(word.reversed())
func isPalindrome(myString:String) -> Bool {
let reverseString = String(myString.characters.reversed())
if(myString != "" && myString == reverseString) {
return true
} else {
return false
}
}
print(isPalindrome("madam"))
I have used the below extension to find whether the number is Palindrome or Not.
extension String {
var isPalindrome: Bool {
return self == String(self.reversed())
}
}
Sometimes having a front end for a recursion can simplify life. I sometimes do this when the arguments which are most convenient to use are not what I want in the user interface.
Would the following meet your needs?
func checkPalindrome(str: String) -> Bool {
func recursiveTest(var charSet: String.CharacterView) -> Bool {
if charSet.count < 2 {
return true
} else {
if charSet.popFirst() != charSet.popLast() {
return false
} else {
return recursiveTest(charSet)
}
}
}
return recursiveTest(str.characters)
}
just add on more condition in if
func checkPalindrome(word: String) -> Bool {
print(word)
if (word == "" || word.characters.count == 1){
return true
}
else {
if word.characters.first == word.characters.last {
return checkPalindrome(word.substringWithRange(word.startIndex.successor() ..< word.endIndex.predecessor()))
} else {
return false
}
}
}
extension StringProtocol where Self: RangeReplaceableCollection {
var letters: Self { filter(\.isLetter) }
var isPalindrome: Bool {
let letters = self.letters
return String(letters.reversed()).caseInsensitiveCompare(letters) == .orderedSame
}
}
"Dammit I'm Mad".isPalindrome // true
"Socorram-me subi no onibus em marrocos".isPalindrome // true
You can also break your string into an array of characters and iterate through them until its half comparing one by one with its counterpart:
func checkPalindrome(_ word: String) -> Bool {
let chars = Array(word.letters.lowercased())
for index in 0..<chars.count/2 {
if chars[index] != chars[chars.count - 1 - index] {
return false
}
}
return true
}
And the recursive version fixing the range issue where can't form a range with endIndex < startIndex:
func checkPalindrome<T: StringProtocol>(_ word: T) -> Bool {
let word = word.lowercased()
.components(separatedBy: .punctuationCharacters).joined()
.components(separatedBy: .whitespacesAndNewlines).joined()
if word == "" || word.count == 1 {
return true
} else {
if word.first == word.last {
let start = word.index(word.startIndex,offsetBy: 1, limitedBy: word.endIndex) ?? word.startIndex
let end = word.index(word.endIndex,offsetBy: -1, limitedBy: word.startIndex) ?? word.endIndex
return checkPalindrome(word[start..<end])
} else {
return false
}
}
}
checkPalindrome("Dammit I'm Mad")
I think if you make an extension to String like this one then it will make your life easier:
extension String {
var length: Int { return characters.count }
subscript(index: Int) -> Character {
return self[startIndex.advancedBy(index)]
}
subscript(range: Range<Int>) -> String {
return self[Range<Index>(start: startIndex.advancedBy(range.startIndex), end: startIndex.advancedBy(range.endIndex))]
}
}
With it in place, you can change your function to this:
func checkPalindrome(word: String) -> Bool {
if word.length < 2 {
return true
}
if word.characters.first != word.characters.last {
return false
}
return checkPalindrome(word[1..<word.length - 1])
}
Quick test:
print(checkPalindrome("aba")) // Prints "true"
print(checkPalindrome("abc")) // Prints "false"
extension String {
func trimmingFirstAndLastCharacters() -> String {
guard let startIndex = index(self.startIndex, offsetBy: 1, limitedBy: self.endIndex) else {
return self
}
guard let endIndex = index(self.endIndex, offsetBy: -1, limitedBy: self.startIndex) else {
return self
}
guard endIndex >= startIndex else {
return self
}
return String(self[startIndex..<endIndex])
}
var isPalindrome: Bool {
guard count > 1 else {
return true
}
return first == last && trimmingFirstAndLastCharacters().isPalindrome
}
}
We first declare a function that removes first and last characters from a string.
Next we declare a computer property which will contain the actual recursive code that checks if a string is palindrome.
If string's size is less than or equal 1 we immediately return true (strings composed by one character like "a" or the empty string "" are considered palindrome), otherwise we check if first and last characters of the string are the same and we recursively call isPalindrome on the current string deprived of the first and last characters.
Convert the string into an Array. When the loop is executed get the first index and compare with the last index.
func palindrome(string: String)-> Bool{
let char = Array(string)
for i in 0..<char.count / 2 {
if char[i] != char[char.count - 1 - i] {
return false
}
}
return true
}
This solution is not recursive, but it is a O(n) pure index based solution without filtering anything and without creating new objects. Non-letter characters are ignored as well.
It uses two indexes and walks outside in from both sides.
I admit that the extension type and property name is stolen from Leo, I apologize. 😉
extension StringProtocol where Self: RangeReplaceableCollection {
var isPalindrome : Bool {
if isEmpty { return false }
if index(after: startIndex) == endIndex { return true }
var forward = startIndex
var backward = endIndex
while forward < backward {
repeat { formIndex(before: &backward) } while !self[backward].isLetter
if self[forward].lowercased() != self[backward].lowercased() { return false }
repeat { formIndex(after: &forward) } while !self[forward].isLetter
}
return true
}
}
Wasn't really thinking of this, but I think I came up with a pretty cool extension, and thought I'd share.
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 test = ["Eye", "Pop", "Noon", "Level", "Radar", "Kayak", "Rotator", "Redivider", "Detartrated", "Tattarrattat", "Aibohphobia", "Eve", "Bob", "Otto", "Anna", "Hannah", "Evil olive", "Mirror rim", "Stack cats", "Doom mood", "Rise to vote sir", "Step on no pets", "Never odd or even", "A nut for a jar of tuna", "No lemon, no melon", "Some men interpret nine memos", "Gateman sees name, garageman sees nametag"]
func checkPalindrome(word: String) -> Bool {
if word.isEmpty { return true }
else {
if word.subString(nil)(1) == word.subString(-1)(nil) {
return checkPalindrome(word.subString(1)(-1))
} else {
return false
}
}
}
for item in test.map({ $0.lowercaseString.stringByReplacingOccurrencesOfString(",", withString: "").stringByReplacingOccurrencesOfString(" ", withString: "") }) {
if !checkPalindrome(item) {
print(item)
}
}
A simple solution in Swift:
func isPalindrome(word: String) -> Bool {
// If no string found, return false
if word.count == 0 { return false }
var index = 0
var characters = Array(word) // make array of characters
while index < characters.count / 2 { // repeat loop only for half length of given string
if characters[index] != characters[(characters.count - 1) - index] {
return false
}
index += 1
}
return true
}
func checkPalindrome(_ inputString: String) -> Bool {
if inputString.count % 2 == 0 {
return false
} else if inputString.count == 1 {
return true
} else {
var stringCount = inputString.count
while stringCount != 1 {
if inputString.first == inputString.last {
stringCount -= 2
} else {
continue
}
}
if stringCount == 1 {
return true
} else {
return false
}
}
}