How to check last 3 character in string contain numbers in Swift? - swift

Question

For a given String instance, I want to check whether the last three characters are numeric characters (0, 1, 2, ..., 9) or not.
For example, the string
let str1 = "SACH092"
should return true for such a query, whereas e.g.
let str2 = "SACHA92"
should return false for the query.
I am using Xcode 7.3.1.

Answer 1

(As pointed out by #NiravD, for pre Swift 3, use where to join parts of multi-clause conditions. For Swift 3, parts of multi-clause conditions are simply joined by ,. For both methods below, both Swift 2.2 and 3 versions are included)
Use pattern matching for numeric characters "0"..."9"
Swift 2.2
extension String {
var lastThreeLettersAreNumbers: Bool {
if case let chars = characters.suffix(3) where chars.count > 2 {
let numbersPattern = Character("0")..."9"
return chars.reduce(true) { $0 && (numbersPattern ~= $1) }
}
return false
}
}
Swift 3
extension String {
var lastThreeLettersAreNumbers: Bool {
if case let chars = characters.suffix(3), chars.count > 2 {
let numbersPattern = Character("0")..."9"
return chars.reduce(true) { $0 && (numbersPattern ~= $1) }
}
return false
}
}
/* example usage, common for both Swift 2.2/3 version */
let str1 = "SACH092"
let str2 = "SACH0B2"
print(str1.lastThreeLettersAreNumbers) // true
print(str2.lastThreeLettersAreNumbers) // false
Make use of nil-return Int by String initializer, with flatMap
You can make use of the fact that the Int by String initializer returns nil for strings that cannot be represented as integers.
Swift 2.2
extension String {
var lastThreeLettersAreNumbers: Bool {
if case let chars = characters.suffix(3) where chars.count > 2 {
return chars.flatMap{Int(String($0))}.count == 3
}
return false
}
}
Swift 3
extension String {
var lastThreeLettersAreNumbers: Bool {
if case let chars = characters.suffix(3), chars.count > 2 {
return chars.flatMap{Int(String($0))}.count == 3
}
return false
}
}
/* example usage, common for both Swift 2.2/3 version */
let str1 = "SACH092"
let str2 = "SACH0B2"
print(str1.lastThreeLettersAreNumbers) // true
print(str2.lastThreeLettersAreNumbers) // false

Answer 2

For getting last 3 characters,
let exampleString = "SACH092"
let last3Char = exampleString.substringFromIndex(exampleString.endIndex.advancedBy(-3))
Check if last3Char contains all Digits,
let badCharacters = NSCharacterSet.decimalDigitCharacterSet().invertedSet
if last3Char.rangeOfCharacterFromSet(badCharacters) == nil {
print("String contains all digits")
} else {
print("String contains non-digit characters")
}

Answer 3

The best way to achieve string control is to use Regular Expressions.
For you :
var str = "SACH092"
let pattern = "^.*[0-9]{3,3}$"
let regexp = try! NSRegularExpression(pattern: pattern, options: [])
let matches = regexp.matches(in: str, options: [], range: NSMakeRange(0, str.characters.count))
print("End with 3 numbers : \(matches.count > 0)")

Answer 4

You can use like this
let s : NSString = "SACH092"
let trimmedString: String = (s as NSString).substringFromIndex(max(s.length-3,0))
print(trimmedString.isNumeric) // return true or false
//Make Extension of String
extension String {
var isNumeric: Bool {
let nums: Set<Character> = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
return Set(self.characters).isSubsetOf(nums)
}
}

Answer 5

To get the last three letters of a string
let oldString = "yourString"
let newString = a.substringFromIndex(a.endIndex.advancedBy(-3))
To check those characters are numbers,
func isNumber(num: String) -> Bool {
let numberCharacters = NSCharacterSet.decimalDigitCharacterSet().invertedSet
return !num.isEmpty && num.rangeOfCharacterFromSet(numberCharacters) == nil
}