Calculator in Swift a method [rangeOfString] - objective-c

Question

I am making calculator in Swift for IOS 8. All operand and operator works very well. Only problem is with "."
For example, 192.168.0.1 is invalid. The "." should only display once. Instead 192.16801 is valid.
See below code -
//Assume user has entered 192.168
//User cannot press "." button again now, "." should not appear again
var display.text 192.168`
if display.text.rangeOfString(".") != nil{
println("exists")
}

Answer 1

import UIKit
class ViewController: UIViewController, UITextFieldDelegate {
#IBOutlet weak var textField: UITextField!
override func viewDidLoad() {
super.viewDidLoad()
self.textField.delegate = self
}
//Textfield delegates
func textField(textField: UITextField, shouldChangeCharactersInRange range: NSRange, replacementString string: String) -> Bool { // return NO to not change text
switch string {
case "0","1","2","3","4","5","6","7","8","9":
return true
case ".":
let array = Array(textField.text)
var decimalCount = 0
for character in array {
if character == "." {
decimalCount++
}
}
if decimalCount == 1 {
return false
} else {
return true
}
default:
let array = Array(string)
if array.count == 0 {
return true
}
return false
}
}
}
the textField functions will return a value of true if you have entered a valid number, it will return a value of false if you input anything else than a number or if you enter a second “.”. If this functions returns false it will not change the value of the text field. Example: if a user (tries to) input 3.14a9 the a will be ignored and the textfield text will be 3.149. If a user inputs 3.45.12 the textfield will have a text value of 3.45.12

Answer 2

I’m not sure this is the question, but if you’re looking for a second “.” you can use the find function with a where for this:
let decimal: Character = "."
if let first = find(display.text, decimal)
where (find(display.text[first.successor()..<display.text.endIndex], decimal) != nil) {
println("second exists")
}
else {
println("none or one")
}

Answer 3

You can use regular expression to validate your textfield whenever its value changes.
Example of a regulator
/*
^ //begin string
([0-9]+)? //optional: has 0 or more digits
(\\.)? //optional: has a point
([0-9]+)? //optional: has 0 or more digits
$ //end string
*/
extension String {
func isValidNumber() -> Bool {
let regex = NSRegularExpression(pattern: "^([0-9]+)?(\\.)?([0-9]+)?$", options: .CaseInsensitive, error: nil)
return regex?.firstMatchInString(self, options: nil, range: NSMakeRange(0, countElements(self))) != nil
}
}
Used in the call-back of your UITextField:
func textField(textField: UITextField, shouldChangeCharactersInRange range: NSRange, replacementString string: String) -> Bool {
return textField.text.isValidNumber()
}

Answer 4

var text = "192.168.2"
var isValid = true
let range = text.rangeOfString(".")
if let range = range {
let subText = text.substringFromIndex(range.endIndex)
let range2 = subText.rangeOfString(".")
if range2 != nil {
isValid = false
}
}
println("isValid: \(isValid)") // isValid: false