单击“标签”显示UIPickerView

Question

我使用UIPickerView为标签选择数据：

class BookingOptionsViewController: UIViewController, UIPickerViewDelegate, UIPickerViewDataSource {

    @IBOutlet weak var lblProfile: UILabel! 

    var data = ["1", "2", "3"]
    var picker = UIPickerView()

    override func viewDidLoad() {
        super.viewDidLoad()        
        picker.delegate = self
        picker.dataSource = self
        let tap = UITapGestureRecognizer(target: self, action: #selector(tap(gestureReconizer:)))
        lblProfile.addGestureRecognizer(tap)
        lblProfile.isUserInteractionEnabled = true
    }

    func tap(gestureReconizer: UITapGestureRecognizer) {
        print("*")
        picker.isHidden = false
    }

func numberOfComponents(in pickerView: UIPickerView) -> Int {
        return 1
    }

    func pickerView(_ pickerView: UIPickerView, numberOfRowsInComponent component: Int) -> Int {
        return data.count
    }    

    func pickerView(_ pickerView: UIPickerView, didSelectRow row: Int, inComponent component: Int) {
        lblProfile.text = data[row]
        self.view.endEditing(true)
    }

    func pickerView(_ pickerView: UIPickerView, titleForRow row: Int, forComponent component: Int) -> String? {
        return data[row]
    }
}

当我点击标签时，我想显示我的UIPickerView，但是UIPickerView不能工作。如何显示UIPickerView？

Answer 1

您从未完成选择器视图的设置。你创造了它。您可以设置它的委托和数据源，但仅此而已。你需要设置它的框架。您需要将其添加到视图控制器的视图(或其他适当的父视图)中。

override func viewDidLoad() {
    super.viewDidLoad()        

    var pickerRect = picker.frame
    pickerRect.origin.x = // some desired value
    pickerRect.origin.y = // some desired value
    picker.frame = pickerRect
    picker.delegate = self
    picker.dataSource = self
    picker.isHidden = true
    view.addSubview(picker)

    let tap = UITapGestureRecognizer(target: self, action: #selector(tap(gestureReconizer:)))
    lblProfile.addGestureRecognizer(tap)
    lblProfile.isUserInteractionEnabled = true
}

Answer 2

您可以创建一个UILabel子类并返回canBecomeFirstResponder true。然后重写inputView以返回所需的pickerView。下面是讨论UILabel doesn't show inputView的一个主题。