带有滑块的UITableView单元格:触摸无法正常工作swift 2 [英] UITableView cell with slider : touch not working correctly swift 2

问题描述

我正在使用UITableView显示多个自定义单元格，而我有一个带有UiSlider的单元格，问题是要移动滑块，我需要长按才能移动它，否则它就不会移动.我试图用UITableView中的倍数滑块来做一个简单的项目，它可以完美地工作.所以我想我需要围绕触摸配置一些东西，但是我不知道是什么.这是我的代码:

I m using a UITableView to display multiple custom cells, and I have one with a UiSlider, the problem is that to move the slider I need to do a long press touch to be able to move it otherwise it don't move. I tried to do a simple project with multiples slider in a UITableView and it works perfectly. So I suppose I need to configure some thing around the touch but I don't know what. Here is my code :

我的视图中有此代码确实已加载手势:

I have this code in my view did load for gesture :

 override func viewDidLoad() {
        super.viewDidLoad()
        let tap: UITapGestureRecognizer = UITapGestureRecognizer(target: self, action: #selector(FormViewController.dismissKeyboard))
        view.addGestureRecognizer(tap)

        // without this line bellow you can't select any cell to display multiple or single choice questions
        tap.cancelsTouchesInView = false
        if  self.navigationController!.respondsToSelector(Selector("interactivePopGestureRecognizer")) {
            self.navigationController!.view.removeGestureRecognizer(navigationController!.interactivePopGestureRecognizer!)
        }

        notificationCenter.addObserver(self, selector: #selector(adjustForKeyboard), name: UIKeyboardWillHideNotification, object: nil)
        notificationCenter.addObserver(self, selector: #selector(adjustForKeyboard), name: UIKeyboardWillChangeFrameNotification, object: nil)

    }
func dismissKeyboard() {
        view.endEditing(true)
    }

对于UITableView cellForRowAtIndexPath方法:

For the UITableView cellForRowAtIndexPath method :

        let cell = tableView.dequeueReusableCellWithIdentifier(CurrentFormTableView.CellIdentifiers.SliderCell, forIndexPath: indexPath) as! SliderCell
        cell.delegate = self
        cell.slider.tag = indexPath.row
        cell.display(block)
        cell.selectionStyle = UITableViewCellSelectionStyle.None
        return cell

这是我的滑块单元格的代码:

And here is the code for my slider cell :

导入UIKit

class SliderCell: UITableViewCell {

    @IBOutlet weak var titleLabel: UILabel!
    @IBOutlet weak var maxLegendLabel: UILabel!
    @IBOutlet weak var minLegendLabel: UILabel!
    @IBOutlet weak var slider: UISlider!
    @IBOutlet weak var answerLabel: UILabel!

    var delegate: QuestionSliderCellDelegate?

    override func awakeFromNib() {
        super.awakeFromNib()
    }

    override func setSelected(selected: Bool, animated: Bool) {
        super.setSelected(selected, animated: animated)
        isFirstResponder()
    }

    @IBAction func slideAction(sender: UISlider) {
        let sliderValue = Int(sender.value)
       print("slider value")
       print(sliderValue)
    }


    func display(block: Block){
        titleLabel.text = block.title
        slider.value = 1.0

    }
}

推荐答案

当您在UITableView中使用其他组件(例如滑块)并且希望获得更好的响应速度时，可以尝试禁用UITableview上的每个延迟.

When you use other components like sliders in UITableView and you want to get better responsiveness, you could try to disable every delay on your UITableview.

以下代码对我来说很有效:

The code below work well for me:

 override func viewDidLoad() {
        super.viewDidLoad()
        yourTableView.delaysContentTouches = false
 }

 func tableView(tableView: UITableView, numberOfRowsInSection section: Int) -> Int {
        for view in tableView.subviews {
            if view is UIScrollView {
                (view as? UIScrollView)!.delaysContentTouches = false
                break
            }
        }
        return numberOfRows
 }

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