How to create an IBInspectable of type enum

IosXcodeSwiftInterface Builder

Ios Problem Overview


enum is not an Interface Builder defined runtime attribute. The following does not show in Interface Builder's Attributes Inspector:

enum StatusShape:Int {
    case Rectangle = 0
    case Triangle = 1
    case Circle = 2
}
@IBInspectable var shape:StatusShape = .Rectangle

From the documentation: 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.

Q: How can I see an enum in Interface Builder's Attributes Inspector?

Ios Solutions


Solution 1 - Ios

Swift 3

@IBInspectable var shape:StatusShape = .Rectangle merely creates a blank entry in Interface Builder:

Not available in IB

Use an adapter, which will acts as a bridge between Swift and Interface Builder.
shapeAdapter is inspectable from IB:

   // IB: use the adapter
   @IBInspectable var shapeAdapter:Int {
        get {
            return self.shape.rawValue
        }
        set( shapeIndex) {
            self.shape = StatusShape(rawValue: shapeIndex) ?? .Rectangle
        }
    }

Available in IB

Unlike the conditional compilation approach (using #if TARGET_INTERFACE_BUILDER), the type of the shape variable does not change with the target, potentially requiring further source code changes to cope with the shape:NSInteger vs. shape:StatusShape variations:

   // Programmatically: use the enum
   var shape:StatusShape = .Rectangle
Complete code
@IBDesignable
class ViewController: UIViewController {

    enum StatusShape:Int {
        case Rectangle
        case Triangle
        case Circle
    }

    // Programmatically: use the enum
    var shape:StatusShape = .Rectangle

    // IB: use the adapter
    @IBInspectable var shapeAdapter:Int {
        get {
            return self.shape.rawValue
        }
        set( shapeIndex) {
            self.shape = StatusShape(rawValue: shapeIndex) ?? .Rectangle
        }
    }
}

► Find this solution on GitHub.

Solution 2 - Ios

Instead of setting your inspectable enums with ints, you could also set them with strings. Although not quite as preferable as a dropdown, at least this option offers some level of readability.

Swift-only Option:

// 1. Set up your enum
enum Shape: String {
    case Rectangle = "rectangle" // lowercase to make it case-insensitive
    case Triangle = "triangle"
    case Circle = "circle"
}


// 2. Then set up a stored property, which will be for use in code
var shape = Shape.Rectangle // default shape


// 3. And another stored property which will only be accessible in IB (because the "unavailable" attribute prevents its use in code)
@available(*, unavailable, message: "This property is reserved for Interface Builder. Use 'shape' instead.")
@IBInspectable var shapeName: String? {
    willSet {
        // Ensure user enters a valid shape while making it lowercase.
        // Ignore input if not valid.
        if let newShape = Shape(rawValue: newValue?.lowercased() ?? "") {
            shape = newShape
        }
    }
}

It is possible to also get this to work with objective-c as well, by adding an initializer to the enum. However, the compiler will only show the "unavailable" error for your IB-only properties in swift code.

Swift Option with Obj-C Compatibility:

@objc enum Shape: Int {
    case None
    case Rectangle
    case Triangle
    case Circle
    
    init(named shapeName: String) {
        switch shapeName.lowercased() {
        case "rectangle": self = .Rectangle
        case "triangle": self = .Triangle
        case "circle": self = .Circle
        default: self = .None
        }
    }
}

var shape = Shape.Rectangle // default shape

@available(*, unavailable, message: "This property is reserved for Interface Builder. Use 'shape' instead.")
@IBInspectable var shapeName: String? {
    willSet {
        if let newShape = Shape(rawValue: newValue?.lowercased() ?? "") {
            shape = newShape
        }
    }
}

Solution 3 - Ios

I can't remember the swift syntax, but this is how I solved it in obj-c

#if TARGET_INTERFACE_BUILDER
@property (nonatomic) IBInspectable NSInteger shape;
#else
@property (nonatomic) StatusShape shape;
#endif

Solution 4 - Ios

For 2020 - @SwiftArchitect answer updated for today:

Here's a typical full example with all of today's syntax

enter image description here

import UIKit

@IBDesignable class ClipLabels: UILabel {
    
    enum Side: Int { case left, right }
    
    var side: Side = .left {
        didSet {
            common()
        }
    }
    
    @available(*, unavailable, message: "IB only")
    @IBInspectable var leftRight01: Int {
        get {
            return self.side.rawValue
        }
        set(index) {
            self.side = Side(rawValue: index) ?? .left
        }
    }
    

and just an example of use ...

switch side {
    case .left:
        textColor = .red
    case .right:
        textColor = .green
    }

For this critical Swift/iOS QA,

• the very old answer of @SwiftArchitect is perfectly correct but

• I've just updated it and added the critical "unavailable" thing, which is now possible in Swift.

Solution 5 - Ios

This is an old thread but useful. I have adapted my answer to swift 4.0 and Xcode 9.0 - Swift 4 has its own little issues with this problem. I am having an @IBInspectable variable with enum type and Xcode 9.0 is not happy, showing me this "Property cannot be marked @IBInspectable because its type cannot be representing in Objective-c"

@Eporediese answers this problem (for swift3) in part; using a property for the storyboard but a straight enum for the rest of the code. Below is a more complete code set that gives you a property to work with in both cases.

enum StatusShape: Int {
  case Rectangle = 0
  case Triangle = 1
  case Circle = 2
}
var _shape:StatusShape = .Rectangle  // this is the backing variable

#if TARGET_INTERFACE_BUILDER
  @IBInspectable var shape: Int {    // using backing variable as a raw int
    
    get { return _shape.rawValue }
    set {
      if _shape.rawValue != newValue {
        _shape.rawValue = newValue
      }
    }
}
#else
var shape: StatusShape {  // using backing variable as a typed enum
  get { return _shape }
  set {
    if _shape != newValue {
      _shape = newValue
    }
  }
}
#endif

Solution 6 - Ios

Swift 3 solution based on [SwiftArchitect][1]

enum StatusShape: Int {
    case rectangle, triangle, circle
}
var statusShape: StatusShape = .rectangle
#if TARGET_INTERFACE_BUILDER
@IBInspectable var statusShapeIB: Int {
    get { 
        return statusShape.rawValue 
    }
    set { 
        guard let statusShape = StatusShape(rawValue: newValue) else { return }
        self.statusShape = statusShape
    }
}   //convenience var, enum not inspectable
#endif

[1]: https://stackoverflow.com/a/27435318/1415703 "SwiftArchitect"

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionSwiftArchitectView Question on Stackoverflow
Solution 1 - IosSwiftArchitectView Answer on Stackoverflow
Solution 2 - IosuɥƃnɐʌuopView Answer on Stackoverflow
Solution 3 - IosJason BobierView Answer on Stackoverflow
Solution 4 - IosFattieView Answer on Stackoverflow
Solution 5 - IosOhmyView Answer on Stackoverflow
Solution 6 - IoslgarboView Answer on Stackoverflow