ios - 在 tableview 中滑动手势

Question

我想在我的手机上添加滑动手势。它工作正常。但问题是当我在第一个单元格上滑动但我的第五个单元格也滑动时。我知道这个索引路径问题。请帮助。但我被困了几个小时。

让 swipeLeft = UISwipeGestureRecognizer(target: self, action: #selector(self.handleLeftSwipe)) swipeLeft.direction = UISwipeGestureRecognizerDirection.left table.addGestureRecognizer(向左滑动)

let swipeLeftd = UISwipeGestureRecognizer(target: self, action: #selector(self.handleLeftSwipes))
swipeLeftd.direction = UISwipeGestureRecognizerDirection.right
table.addGestureRecognizer(swipeLeftd)

@objc func handleLeftSwipe(sender: UITapGestureRecognizer) {
    print("rigt called")

    let location = sender.location(in: self.table)
    let indexPath = self.table.indexPathForRow(at: location)
    let cell = self.table.cellForRow(at: indexPath!) as! TableViewCell
    print("swipe")
    cell.myView.frame.origin.x = -94
}

@objc func handleLeftSwipes(sender: UITapGestureRecognizer) {
    print("labelSwipedLeft called")
    let location = sender.location(in: self.table)
    let indexPath = self.table.indexPathForRow(at: location)
    let cell = self.table.cellForRow(at: indexPath!) as! TableViewCell
    print("swipe")
    cell.myView.frame.origin.x =  80
}

func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
    let identifier = "TableViewCell"
    var cell: TableViewCell! = tableView.dequeueReusableCell(withIdentifier: identifier) as? TableViewCell
    if cell == nil {
        var nib : Array = Bundle.main.loadNibNamed("TableViewCell",owner: self,options: nil)!
        cell = nib[2] as? TableViewCell
    }

    return cell!
}

Answer 1

这样做的原因是重复使用您的细胞。我假设您没有在 prepareForReuse() 中设置任何内容。那么，您的问题在 cell.myView.frame.origin.x = 80 行，对吗？只需在 prepareForReuse 或 cellForRowAt indexPath 中将其设置为默认值即可。