Related
Like my question title.
If I get string 00001, I want to +1 and return 00002.
If I get string 00009, I want to +1 and return 00010.
....
If I get string 89999, I want to +1 and return 90000.
How to make this function looks better and easier? Thanks.
func createFloatNumberString(str: String){
var idArray: [String] = [] //[0,0,0,0,1] [0,0,0,1,0] [8,9,9,9,9]
str.map { (char) in
idArray.append(char.description)
}
idArray[4] = String(Int(idArray[4])!+1)
if idArray[4] == "10" {
idArray[4] = "0"
idArray[3] = String(Int(idArray[3])!+1)
if idArray[3] == "10" {
idArray[3] = "0"
idArray[2] = String(Int(idArray[2])!+1)
if idArray[2] == "10" {
idArray[2] = "0"
idArray[1] = String(Int(idArray[1])!+1)
if idArray[1] == "10" {
idArray[1] = "0"
idArray[0] = String(Int(idArray[0])!+1)
}
}
}
}
}
You need to convert the String to an Int, increment it, then use a formatted String as the return value.
func incrementFormattedNumericString(numericString:String,increment by:Int=1) -> String? {
guard let numericValue = Int(numericString) else { return nil }
return String(format: "%.5ld", numericValue+by)
}
If you are sure that the numericString passed into the function will always be a valid number, you can force unwrap the conversion to Int and return a non-Optional value.
Very simple solution:
As counter use an Int and this extension
extension Int {
var fiveDigitStringRepresentation : String {
return String(format: "%05ld", self)
}
}
123.fiveDigitStringRepresentation // "00123"
Or – a bit more sophisticated – use a struct
struct Count {
var counter = 0
mutating func increment() { counter += 1 }
var fiveDigitStringRepresentation : String {
return String(format: "%05ld", counter)
}
}
var count = Count()
count.increment()
count.increment()
print(count.fiveDigitStringRepresentation) // "00002"
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
}
}
}
Is there a way to check strings for nil and "" in Swift? In Rails, I can use blank() to check.
I currently have this, but it seems overkill:
if stringA? != nil {
if !stringA!.isEmpty {
...blah blah
}
}
If you're dealing with optional Strings, this works:
(string ?? "").isEmpty
The ?? nil coalescing operator returns the left side if it's non-nil, otherwise it returns the right side.
You can also use it like this to return a default value:
(string ?? "").isEmpty ? "Default" : string!
You could perhaps use the if-let-where clause:
Swift 3:
if let string = string, !string.isEmpty {
/* string is not blank */
}
Swift 2:
if let string = string where !string.isEmpty {
/* string is not blank */
}
With Swift 5, you can implement an Optional extension for String type with a boolean property that returns if an optional string is empty or has no value:
extension Optional where Wrapped == String {
var isEmptyOrNil: Bool {
return self?.isEmpty ?? true
}
}
However, String implements isEmpty property by conforming to protocol Collection. Therefore we can replace the previous code's generic constraint (Wrapped == String) with a broader one (Wrapped: Collection) so that Array, Dictionary and Set also benefit our new isEmptyOrNil property:
extension Optional where Wrapped: Collection {
var isEmptyOrNil: Bool {
return self?.isEmpty ?? true
}
}
Usage with Strings:
let optionalString: String? = nil
print(optionalString.isEmptyOrNil) // prints: true
let optionalString: String? = ""
print(optionalString.isEmptyOrNil) // prints: true
let optionalString: String? = "Hello"
print(optionalString.isEmptyOrNil) // prints: false
Usage with Arrays:
let optionalArray: Array<Int>? = nil
print(optionalArray.isEmptyOrNil) // prints: true
let optionalArray: Array<Int>? = []
print(optionalArray.isEmptyOrNil) // prints: true
let optionalArray: Array<Int>? = [10, 22, 3]
print(optionalArray.isEmptyOrNil) // prints: false
Sources:
swiftbysundell.com - Extending optionals in Swift
objc.io - Swift Tip: Non-Empty Collections
Using the guard statement
I was using Swift for a while before I learned about the guard statement. Now I am a big fan. It is used similarly to the if statement, but it allows for early return and just makes for much cleaner code in general.
To use guard when checking to make sure that a string is neither nil nor empty, you can do the following:
let myOptionalString: String? = nil
guard let myString = myOptionalString, !myString.isEmpty else {
print("String is nil or empty.")
return // or break, continue, throw
}
/// myString is neither nil nor empty (if this point is reached)
print(myString)
This unwraps the optional string and checks that it isn't empty all at once. If it is nil (or empty), then you return from your function (or loop) immediately and everything after it is ignored. But if the guard statement passes, then you can safely use your unwrapped string.
See Also
Statements documentation
The Guard Statement in Swift 2
If you are using Swift 2, here is an example my colleague came up with, which adds isNilOrEmpty property on optional Strings:
protocol OptionalString {}
extension String: OptionalString {}
extension Optional where Wrapped: OptionalString {
var isNilOrEmpty: Bool {
return ((self as? String) ?? "").isEmpty
}
}
You can then use isNilOrEmpty on the optional string itself
func testNilOrEmpty() {
let nilString:String? = nil
XCTAssertTrue(nilString.isNilOrEmpty)
let emptyString:String? = ""
XCTAssertTrue(emptyString.isNilOrEmpty)
let someText:String? = "lorem"
XCTAssertFalse(someText.isNilOrEmpty)
}
var str: String? = nil
if str?.isEmpty ?? true {
print("str is nil or empty")
}
str = ""
if str?.isEmpty ?? true {
print("str is nil or empty")
}
I know there are a lot of answers to this question, but none of them seems to be as convenient as this (in my opinion) to validate UITextField data, which is one of the most common cases for using it:
extension Optional where Wrapped == String {
var isNilOrEmpty: Bool {
return self?.trimmingCharacters(in: .whitespaces).isEmpty ?? true
}
}
You can just use
textField.text.isNilOrEmpty
You can also skip the .trimmingCharacters(in:.whitespaces) if you don't consider whitespaces as an empty string or use it for more complex input tests like
var isValidInput: Bool {
return !isNilOrEmpty && self!.trimmingCharacters(in: .whitespaces).characters.count >= MIN_CHARS
}
If you want to access the string as a non-optional, you should use Ryan's Answer, but if you only care about the non-emptiness of the string, my preferred shorthand for this is
if stringA?.isEmpty == false {
...blah blah
}
Since == works fine with optional booleans, I think this leaves the code readable without obscuring the original intention.
If you want to check the opposite: if the string is nil or "", I prefer to check both cases explicitly to show the correct intention:
if stringA == nil || stringA?.isEmpty == true {
...blah blah
}
I would recommend.
if stringA.map(isEmpty) == false {
println("blah blah")
}
map applies the function argument if the optional is .Some.
The playground capture also shows another possibility with the new Swift 1.2 if let optional binding.
SWIFT 3
extension Optional where Wrapped == String {
/// Checks to see whether the optional string is nil or empty ("")
public var isNilOrEmpty: Bool {
if let text = self, !text.isEmpty { return false }
return true
}
}
Use like this on optional string:
if myString.isNilOrEmpty { print("Crap, how'd this happen?") }
Swift 3
For check Empty String best way
if !string.isEmpty{
// do stuff
}
Swift 3 solution
Use the optional unwrapped value and check against the boolean.
if (string?.isempty == true) {
// Perform action
}
You should do something like this:
if !(string?.isEmpty ?? true) { //Not nil nor empty }
Nil coalescing operator checks if the optional is not nil, in case it is not nil it then checks its property, in this case isEmpty. Because this optional can be nil you provide a default value which will be used when your optional is nil.
Based on this Medium post, with a little tweak for Swift 5, I got to this code that worked.
if let stringA, !stringA.isEmpty {
...blah blah
}
Although I understand the benefits of creating an extension, I thought it might help someone needing just for a small component / package.
You can create your own custom function, if that is something you expect to do a lot.
func isBlank (optionalString :String?) -> Bool {
if let string = optionalString {
return string.isEmpty
} else {
return true
}
}
var optionalString :String? = nil
if isBlank(optionalString) {
println("here")
}
else {
println("there")
}
Create a String class extension:
extension String
{ // returns false if passed string is nil or empty
static func isNilOrEmpty(_ string:String?) -> Bool
{ if string == nil { return true }
return string!.isEmpty
}
}// extension: String
Notice this will return TRUE if the string contains one or more blanks. To treat blank string as "empty", use...
return string!.trimmingCharacters(in: CharacterSet.whitespaces).isEmpty
... instead. This requires Foundation.
Use it thus...
if String.isNilOrEmpty("hello world") == true
{ print("it's a string!")
}
Swift 3
This works well to check if the string is really empty. Because isEmpty returns true when there's a whitespace.
extension String {
func isEmptyAndContainsNoWhitespace() -> Bool {
guard self.isEmpty, self.trimmingCharacters(in: .whitespaces).isEmpty
else {
return false
}
return true
}
}
Examples:
let myString = "My String"
myString.isEmptyAndContainsNoWhitespace() // returns false
let myString = ""
myString.isEmptyAndContainsNoWhitespace() // returns true
let myString = " "
myString.isEmptyAndContainsNoWhitespace() // returns false
Using isEmpty
"Hello".isEmpty // false
"".isEmpty // true
Using allSatisfy
extension String {
var isBlank: Bool {
return allSatisfy({ $0.isWhitespace })
}
}
"Hello".isBlank // false
"".isBlank // true
Using optional String
extension Optional where Wrapped == String {
var isBlank: Bool {
return self?.isBlank ?? true
}
}
var title: String? = nil
title.isBlank // true
title = ""
title.isBlank // true
Reference : https://useyourloaf.com/blog/empty-strings-in-swift/
This is a general solution for all types that conform to the Collection protocol, which includes String:
extension Optional where Wrapped: Collection {
var isNilOrEmpty: Bool {
self?.isEmpty ?? true
}
}
When dealing with passing values from local db to server and vice versa, I was having too much trouble with ?'s and !'s and what not.
So I made a Swift3.0 utility to handle null cases and i can almost totally avoid ?'s and !'s in the code.
func str(_ string: String?) -> String {
return (string != nil ? string! : "")
}
Ex:-
Before :
let myDictionary: [String: String] =
["title": (dbObject?.title != nil ? dbObject?.title! : "")]
After :
let myDictionary: [String: String] =
["title": str(dbObject.title)]
and when its required to check for a valid string,
if !str(dbObject.title).isEmpty {
//do stuff
}
This saved me having to go through the trouble of adding and removing numerous ?'s and !'s after writing code that reasonably make sense.
Use the ternary operator (also known as the conditional operator, C++ forever!):
if stringA != nil ? stringA!.isEmpty == false : false { /* ... */ }
The stringA! force-unwrapping happens only when stringA != nil, so it is safe. The == false verbosity is somewhat more readable than yet another exclamation mark in !(stringA!.isEmpty).
I personally prefer a slightly different form:
if stringA == nil ? false : stringA!.isEmpty == false { /* ... */ }
In the statement above, it is immediately very clear that the entire if block does not execute when a variable is nil.
helpful when getting value from UITextField and checking for nil & empty string
#IBOutlet weak var myTextField: UITextField!
Heres your function (when you tap on a button) that gets string from UITextField and does some other stuff
#IBAction func getStringFrom_myTextField(_ sender: Any) {
guard let string = myTextField.text, !(myTextField.text?.isEmpty)! else { return }
//use "string" to do your stuff.
}
This will take care of nil value as well as empty string.
It worked perfectly well for me.
Swift 5.6 - Xcode 13
extension Optional where Wrapped: Collection {
var isEmptyOrNil: Bool {
guard let self = self else { return true }
return self.isEmpty
}
}
Usage:
var name: String?
if name.isEmptyOrNil {
///true
}
name = "John Peter"
guard !name.isEmptyOrNil else { return }
/// Name is not empty
you can use this func
class func stringIsNilOrEmpty(aString: String) -> Bool { return (aString).isEmpty }
I have an array ["abc", "94761178","790"]
I want to iterate each and check is a String or an Int?
How to check it?
How to convert "123" to integer 123?
Here is a small Swift version using String extension :
Swift 3/Swift 4 :
extension String {
var isNumber: Bool {
return !isEmpty && rangeOfCharacter(from: CharacterSet.decimalDigits.inverted) == nil
}
}
Swift 2 :
extension String {
var isNumber : Bool {
get{
return !self.isEmpty && self.rangeOfCharacterFromSet(NSCharacterSet.decimalDigitCharacterSet().invertedSet) == nil
}
}
}
Edit Swift 2.2:
In swift 2.2 use Int(yourArray[1])
var yourArray = ["abc", "94761178","790"]
var num = Int(yourArray[1])
if num != nil {
println("Valid Integer")
}
else {
println("Not Valid Integer")
}
It will show you that string is valid integer and num contains valid Int.You can do your calculation with num.
From docs:
If the string represents an integer that fits into an Int, returns the
corresponding integer.This accepts strings that match the regular
expression "[-+]?[0-9]+" only.
Be aware that checking a string/number using the Int initializer has limits. Specifically, a max value of 2^32-1 or 4294967295. This can lead to problems, as a phone number of 8005551234 will fail the Int(8005551234) check despite being a valid number.
A much safer approach is to use NSCharacterSet to check for any characters matching the decimal set in the range of the string.
let number = "8005551234"
let numberCharacters = NSCharacterSet.decimalDigitCharacterSet().invertedSet
if !number.isEmpty && number.rangeOfCharacterFromSet(numberCharacters) == nil {
// string is a valid number
} else {
// string contained non-digit characters
}
Additionally, it could be useful to add this to a String extension.
public extension String {
func isNumber() -> Bool {
let numberCharacters = NSCharacterSet.decimalDigitCharacterSet().invertedSet
return !self.isEmpty && self.rangeOfCharacterFromSet(numberCharacters) == nil
}
}
I think the nicest solution is:
extension String {
var isNumeric : Bool {
return Double(self) != nil
}
}
Starting from Swift 2, String.toInt() was removed.
A new Int Initializer was being introduced: Int(str: String)
for target in ["abc", "94761178","790"]
{
if let number = Int(target)
{
print("value: \(target) is a valid number. add one to get :\(number+1)!")
}
else
{
print("value: \(target) is not a valid number.")
}
}
Swift 3, 4
extension String {
var isNumber: Bool {
let characters = CharacterSet.decimalDigits.inverted
return !self.isEmpty && rangeOfCharacter(from: characters) == nil
}
}
Simple solution like this:
extension String {
public var isNumber: Bool {
return !isEmpty && rangeOfCharacter(from: CharacterSet.decimalDigits.inverted) == nil
}
}
I think using NumberFormatter is an easy way:
(Swift 5)
import Foundation
extension String {
private static let numberFormatter = NumberFormatter()
var isNumeric : Bool {
Self.numberFormatter.number(from: self) != nil
}
}
The correct way is to use the toInt() method of String, and an optional binding to determine whether the conversion succeeded or not. So your loop would look like:
let myArray = ["abc", "94761178","790"]
for val in myArray {
if let intValue = val.toInt() {
// It's an int
println(intValue)
} else {
// It's not an int
println(val)
}
}
The toInt() method returns an Int?, so an optional Int, which is nil if the string cannot be converted ton an integer, or an Int value (wrapped in the optional) if the conversion succeeds.
The method documentation (shown using CMD+click on toInt in Xcode) says:
If the string represents an integer that fits into an Int, returns the corresponding integer. This accepts strings that match the regular expression "[-+]?[0-9]+" only.
This way works also with strings with mixed numbers:
public extension String {
func isNumber() -> Bool {
return !self.isEmpty && self.rangeOfCharacter(from: CharacterSet.decimalDigits) != nil && self.rangeOfCharacter(from: CharacterSet.letters) == nil
}}
So u get something like this:
Swift 3.0 version
func isNumber(stringToTest : String) -> Bool {
let numberCharacters = CharacterSet.decimalDigits.inverted
return !s.isEmpty && s.rangeOfCharacter(from:numberCharacters) == nil
}
If you want to accept a more fine-grained approach (i.e. accept a number like 4.5 or 3e10), you proceed like this:
func isNumber(val: String) -> Bool
{
var result: Bool = false
let parseDotComNumberCharacterSet = NSMutableCharacterSet.decimalDigitCharacterSet()
parseDotComNumberCharacterSet.formUnionWithCharacterSet(NSCharacterSet(charactersInString: ".e"))
let noNumberCharacters = parseDotComNumberCharacterSet.invertedSet
if let v = val
{
result = !v.isEmpty && v.rangeOfCharacterFromSet(noNumberCharacters) == nil
}
return result
}
For even better resolution, you might draw on regular expression..
Xcode 8 and Swift 3.0
We can also check :
//MARK: - NUMERIC DIGITS
class func isString10Digits(ten_digits: String) -> Bool{
if !ten_digits.isEmpty {
let numberCharacters = NSCharacterSet.decimalDigits.inverted
return !ten_digits.isEmpty && ten_digits.rangeOfCharacter(from: numberCharacters) == nil
}
return false
}
This code works for me for Swift 3/4
func isNumber(textField: UITextField) -> Bool {
let allowedCharacters = CharacterSet.decimalDigits
let characterSet = CharacterSet(charactersIn: textField.text!)
return allowedCharacters.isSuperset(of: characterSet)
// return true
}
You can use this for integers of any length.
func getIntegerStrings(from givenStrings: [String]) -> [String]
{
var integerStrings = [String]()
for string in givenStrings
{
let isValidInteger = isInteger(givenString: string)
if isValidInteger { integerStrings.append(string) }
}
return integerStrings
}
func isInteger(givenString: String) -> Bool
{
var answer = true
givenString.forEach { answer = ("0"..."9").contains($0) && answer }
return answer
}
func getIntegers(from integerStrings: [String]) -> [Int]
{
let integers = integerStrings.compactMap { Int($0) }
return integers
}
let strings = ["abc", "94761178", "790", "18446744073709551615000000"]
let integerStrings = getIntegerStrings(from: strings)
let integers = getIntegers(from: integerStrings)
print(integerStrings) // ["94761178", "790", "18446744073709551615000000"]
print(integers) // [94761178, 790]
However, as pointed out by #Can, you can get the integer value for the number only up to 2^31 - 1 (signed integer limit on 32-bit arch). For the larger value, however, you will still get the string representation.
This code will return an array of converted integers:
["abc", "94761178","790"].map(Int.init) // returns [ nil, 94761178, 790 ]
OR
["abc", "94761178","790"].map { Int($0) ?? 0 } // returns [ 0, 94761178, 790 ]
Get the following isInteger() function from the below stackoverflow post posted by corsiKa:
Determine if a String is an Integer in Java
And I think this is what you want to do (where nameOfArray is the array you want to pass)
void convertStrArrayToIntArray( int[] integerArray ) {
for (int i = 0; i < nameOfArray.length(); i++) {
if (!isInteger(nameOfArray[i])) {
integerArray[i] = nameOfArray[i].toString();
}
}
}