在 Swift 中访问 UISegementedControl

Question

我一直在使用 UISegmentedControl 并希望根据需要对其进行管理。我在故事板上创建了它，连接了出口和动作。它可以很好地可视化，并且操作 indexchanged 上的回调有效。我想在可视化之前对此进行一些其他管理，所以我被困在如何正确访问我在情节提要中设置的 mysegmentedControl 上。似乎是一个愚蠢的问题，但我找不到当前可以涉及的 UISegmentedControl 的 Swift 示例。

class ViewController: UIViewController, MKMapViewDelegate {
    ...
    //playing around with a Segmented Control
    @IBOutlet var mysegmentedControl : UISegmentedControl

    // Any time the button is clicked in the Segmented Control the index changes so we catch it to do something.

    @IBAction func indexChanged(sender : UISegmentedControl) {
        // This all works fine and it prints out the value of 3 on any click 
        println("# of Segments = \(sender.numberOfSegments)")

        switch sender.selectedSegmentIndex {
        case 0:
            println("first segement clicked")
        case 1:
            println("second segment clicked")
        case 2:
            println("third segemnet clicked")
        default:
            break;
        }  //Switch
    } // indexChanged for the Segmented Control

    override func viewDidLoad() {
        super.viewDidLoad()

         //  1.  How do I get proper access to the mysegmentedControl that was created in the storyboard and is already displayed.
         //      The indexChanged function gets called fine on the display via sender getting passed to it.
         //      I equate this to getting the pointer to it in C so I leverage it elsewhere
         //  2.  Guessing I don't need to init a new UISegmentedControl or do I?

         // This compiles but crashes on run with: fatal error: Can't unwrap 
         println("# of Segments = \(mysegmentedControl.numberOfSegments)")

    } // viewDidLoad    

} // ViewController

Answer 1

我能够成功编译和运行您的代码。您的问题不是您的分段 ctrl，可能您的可选对象之一是 nil，因此您无法打开它。看看你的可选对象。

另外，试试 @IBOutlet var mysegmentedControl : UISegmentedControl?

这绝对可以解决您的问题

Answer 2

这是我用过的，对我来说很好用 -

@IBOutlet var mysegmentedControl : UISegmentedControl?

override func didReceiveMemoryWarning() {
    super.didReceiveMemoryWarning()
    // Dispose of any resources that can be recreated.
}

@IBAction func indexChanged(sender : UISegmentedControl) {
    // This all works fine and it prints out the value of 3 on any click
    println("# of Segments = \(sender.numberOfSegments)")

    switch sender.selectedSegmentIndex {
    case 0:
        println("first segement clicked")
    case 1:
        println("second segment clicked")
    case 2:
        println("third segemnet clicked")
    default:
        break;
    }  //Switch
} // indexChanged for the Segmented Control

override func viewDidLoad() {
    super.viewDidLoad()
    println("# of Segments = \(mysegmentedControl?.numberOfSegments)")

}

Answer 3

这就是应如何声明 IBOutlet。事实上，Xcode 6 Beta 4 会自动为所有 IBOutlets 执行此操作。这 ！将变量标记为隐式展开的可选：

@IBOutlet var mysegmentedControl : UISegmentedControl!

https://developer.apple.com/library/prerelease/ios/documentation/Swift/Conceptual/Swift_Programming_Language/Types.html

阅读“隐式展开的可选类型”部分。

Answer 4

请找到以下代码 sn-p 用于使用 swift 访问 uisegmentedcontrol。

   @IBOutlet weak var segmentedControl: UISegmentedControl!

    @IBOutlet weak var textLabel: UILabel!

    override func viewDidLoad() {
        super.viewDidLoad()
        // Do any additional setup after loading the view, typically from a nib.
        textLabel.text = "First Segment Selected";
    }

    override func didReceiveMemoryWarning() {
        super.didReceiveMemoryWarning()
        // Dispose of any resources that can be recreated.
    }


    @IBAction func segmentedControlAction(sender: AnyObject) {

        if(segmentedControl.selectedSegmentIndex == 0)
        {
            textLabel.text = "First Segment Selected";
        }
        else if(segmentedControl.selectedSegmentIndex == 1)
        {
            textLabel.text = "Second Segment Selected";
        }
        else if(segmentedControl.selectedSegmentIndex == 2)
        {
            textLabel.text = "Third Segment Selected";
        }
    }

如果需要详细说明，请参考以下链接。

http://sourcefreeze.com/uisegmentedcontrol-example-using-swift-in-ios/