如何设置IBInspectable Int的最大限制 [英] How to set a max limit for an IBInspectable Int

问题描述

我在Swift中使用IBInspectable Int来选择4种形状（0-3），但在故事板编辑器中可以设置大于3且小于0的值，这会阻止IBDesignable系统工作。

I am using an IBInspectable Int in Swift to choose between 4 four shapes (0-3), however it is possible in the storyboard editor to set a value greater than 3 and less than 0, which stops the IBDesignable system working.

是否可以设置故事板编辑器中可以设置的最小值和最大值？

Is it possible to set a min and max limit of what values can be set in the storyboard editor?

let SHAPE_CROSS = 0
let SHAPE_SQUARE = 1
let SHAPE_CIRCLE = 2
let SHAPE_TRIANGLE = 3

@IBInspectable var shapeType: Int = 0
@IBInspectable var shapeSize: CGFloat = 100.0
@IBInspectable var shapeColor: UIColor?

推荐答案

没有办法限制用户输入的内容故事板。但是，您可以使用计算属性阻止无效值的存储：

There's no way to limit what a user can input in Storyboard. However, you could prevent invalid values from being stored using a computed property:

  @IBInspectable var shapeType: Int {
    set(newValue) {
      internalShapeType = min(newValue, 3)
    }
    get {
      return internalShapeType
    }
  }

  var internalShapeType: Int = 0

然后你也可以使用枚举而不是常量来在内部表示不同的形状类型。

Then you could also use an enum instead of constants to represent your different shape types internally.

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