<分区>
<分区>
我有一个表格 View ，单击标签时我想使用弹出窗口方法显示 Storyboard 中的 UIViewController。我在点击识别器选择器中有以下代码
func setupItemNameTapRecognizer(_ label:UILabel) {
label.isUserInteractionEnabled = true
let tapRecog = UITapGestureRecognizer(target: self, action: #selector(self.actionItemNameTap(_:)))
label.addGestureRecognizer(tapRecog)
}
func actionItemNameTap(_ sender:UIView) {
print("item tap")
let indexPath = IndexPath(row: sender.tag, section: 0)
let cell = tableView.cellForRow(at:indexPath )
self.showPopOverBox(cell: cell!)
}
和 CellForRowAt 方法中的以下代码
let cell = tableView.dequeueReusableCell(withIdentifier: "ItemContentCell", for: indexPath) as! ItemContentCell
setupItemNameTapRecognizer(cell.itemName)
cell.itemName.tag = indexPath.row
return cell
每当我单击标签时，都会抛出以下错误，但没有发现问题
[UITapGestureRecognizer tag]: unrecognized selector sent to instance 0x7fdc1867ee90 2017-05-23 17:36:23.871 InvoiceMaster[71236:14670269] *** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '-[UITapGestureRecognizer tag]: unrecognized
最佳答案
jsut change this method like below
func actionItemNameTap(_ sender: UITapGestureRecognizer) {
// let view = sender.view;
// print("\(view?.tag)")
print("item tap")
let indexPath = IndexPath(row: (sender.view?.tag)!, section: 0)
let cell = tableView.cellForRow(at:indexPath )
self.showPopOverBox(cell: cell!)
}
关于ios - 无法从 UITableViewCell 显示弹出窗口，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/44134398/