Want to achieve the Keyboard selection through #IBInspectable - ios

Question

I wanted to create a keyboard selection from #IBInspectabale
How to achieve this
I am creating a view, In which I insert an ImageView and a TextField,
Now I am creating this custom view class as #IBDesignable and created the #IBInspectable elements.
I successfully create side image and placeholder elements but now I am trying to create the keyboard type but facing issues.
code snipped : `import UIKit
#IBDesignable
class CustomTextField: UIView,UITextFieldDelegate {
//custom view from the XIB file
var view: UIView!
#IBOutlet weak var textField: UITextField!
#IBOutlet weak var imageView: UIImageView!
override init(frame: CGRect) {
super.init(frame: frame)
loadViewFromNib ()
}
required init?(coder aDecoder: NSCoder) {
super.init(coder: aDecoder)
loadViewFromNib ()
}
func loadViewFromNib() {
let bundle = Bundle(for: type(of: self))
let nib = UINib(nibName: "CustomTextField", bundle: bundle)
let view = nib.instantiate(withOwner: self, options: nil)[0] as! UIView
view.frame = bounds
view.autoresizingMask = [.flexibleWidth, .flexibleHeight]
self.addSubview(view);
}
#IBInspectable var sideImage: UIImage? {
get {
return imageView.image
}
set(sideImage) {
imageView.image = sideImage
}
}
#IBInspectable var placeHolderText: String? {
get {
return textField.placeholder
}
set(placeHolderText) {
textField.placeholder = placeHolderText
}
}'
all the above working fine,
but following is not working for me:
#IBInspectable var keyboard: UIKeyboardType? {
get{
return UIKeyboardType(rawValue: textField.keyboardType.rawValue)
}
set(keyboard){
textField.keyboardType = keyboard!
}
}
}
I tried it by creating enum but it does not give any result for me.

Answer 1

First of all thanks to everyone.
My problem is solved without doing any extra effort of creating enum and all.
I used the apple predefined UIKeyboardType enum.
Just write the following code:
#IBInspectable var keyboard:Int{
get{
return self.textField.keyboardType.rawValue
}
set(keyboardIndex){
self.textField.keyboardType = UIKeyboardType.init(rawValue: keyboardIndex)!
}
}
And it will show the Keyboard in Interface builder, and you can set 0,1,2... value for your keyboard type.
where the 0,1,2 represent as follows:
0: default // Default type for the current input method.
1: asciiCapable // Displays a keyboard which can enter ASCII characters
2: numbersAndPunctuation // Numbers and assorted punctuation.
3: URL // A type optimized for URL entry (shows . / .com prominently).
4: numberPad // A number pad with locale-appropriate digits (0-9, ۰-۹, ०-९, etc.). Suitable for PIN entry.
5: phonePad // A phone pad (1-9, *, 0, #, with letters under the numbers).
6: namePhonePad // A type optimized for entering a person's name or phone number.
7: emailAddress // A type optimized for multiple email address entry (shows space # . prominently).
8: decimalPad // A number pad with a decimal point.
9: twitter // A type optimized for twitter text entry (easy access to # #)

Answer 2

It is not possible to use enum types for #IBInspectable vars. You have to set your var as a String or Int.
From Apple's docs:
You can attach the IBInspectable attribute to any property in a class
declaration, class extension, or category for any type that’s
supported by the Interface Builder defined runtime attributes:
boolean, integer or floating point number, string, localized string,
rectangle, point, size, color, range, and nil.