How to check if a string contains an int? -Swift - string

Question

I need to know if a string contains an Int to be sure that a name the user entered is a valid full name,
for that I need to either make the user type only chars, or valid that there are no ints in the string the user entered.
Thanks for all the help.

Answer 1

You can use Foundation methods with Swift strings, and that's what you should do here. NSString has built in methods that use NSCharacterSet to check if certain types of characters are present. This translates nicely to Swift:
var str = "Hello, playground1"
let decimalCharacters = CharacterSet.decimalDigits
let decimalRange = str.rangeOfCharacter(from: decimalCharacters)
if decimalRange != nil {
print("Numbers found")
}
If you're interested in restricting what can be typed, you should implement UITextFieldDelegate and the method textField(_:shouldChangeCharactersIn:replacementString:) to prevent people from typing those characters in the first place.

Answer 2

Simple Swift 4 version using rangeOfCharacter method from String class:
let numbersRange = stringValue.rangeOfCharacter(from: .decimalDigits)
let hasNumbers = (numbersRange != nil)

Answer 3

This method is what i use now for checking if a string contains a number
func doStringContainsNumber( _string : String) -> Bool{
let numberRegEx = ".*[0-9]+.*"
let testCase = NSPredicate(format:"SELF MATCHES %#", numberRegEx)
let containsNumber = testCase.evaluateWithObject(_string)
return containsNumber
}
If your string Contains a number it will return true else false. Hope it helps

Answer 4

//Swift 3.0 to check if String contains numbers (decimal digits):
let someString = "string 1"
let numberCharacters = NSCharacterSet.decimalDigits
if someString.rangeOfCharacter(from: numberCharacters) != nil
{ print("String contains numbers")}
else if someString.rangeOfCharacter(from: numberCharacters) == nil
{ print("String doesn't contains numbers")}

Answer 5

//A function that checks if a string has any numbers
func stringHasNumber(_ string:String) -> Bool {
for character in string{
if character.isNumber{
return true
}
}
return false
}
/// Check stringHasNumber function
stringHasNumber("mhhhldiddld")
stringHasNumber("kjkdjd99900")

Answer 6

if (ContainsNumbers(str).count > 0)
{
// Your string contains at least one number 0-9
}
func ContainsNumbers(s: String) -> [Character]
{
return s.characters.filter { ("0"..."9").contains($0)}
}

Answer 7

Swift 2.3. version working.
extension String
{
func containsNumbers() -> Bool
{
let numberRegEx = ".*[0-9]+.*"
let testCase = NSPredicate(format:"SELF MATCHES %#", numberRegEx)
return testCase.evaluateWithObject(self)
}
}
Usage:
//guard let firstname = textField.text else { return }
let testStr1 = "lalalala"
let testStr2 = "1lalalala"
let testStr3 = "lal2lsd2l"
print("Test 1 = \(testStr1.containsNumbers())\nTest 2 = \(testStr2.containsNumbers())\nTest 3 = \(testStr3.containsNumbers())\n")

Answer 8

You need to trick Swift into using Regex by wrapping up its nsRegularExpression
class Regex {
let internalExpression: NSRegularExpression
let pattern: String
init(_ pattern: String) {
self.pattern = pattern
var error: NSError?
self.internalExpression = NSRegularExpression(pattern: pattern, options: .CaseInsensitive, error: &error)
}
func test(input: String) -> Bool {
let matches = self.internalExpression.matchesInString(input, options: nil, range:NSMakeRange(0, countElements(input)))
return matches.count > 0
}
}
if Regex("\\d/").test("John 2 Smith") {
println("has a number in the name")
}
I got these from http://benscheirman.com/2014/06/regex-in-swift/

Answer 9

let numericCharSet = CharacterSet.init(charactersIn: "1234567890")
let newCharSet = CharacterSet.init(charactersIn: "~`##$%^&*(){}[]<>?")
let sentence = "Tes#ting4 #Charact2er1Seqt"
if sentence.rangeOfCharacter(from: numericCharSet) != nil {
print("Yes It,Have a Numeric")
let removedSpl = sentence.components(separatedBy: newCharSet).joined()
print(sentence.components(separatedBy: newCharSet).joined())
print(removedSpl.components(separatedBy: numericCharSet).joined())
}
else {
print("No")
}