我试图在单击放置在表格视图单元格中的按钮时显示一个弹出菜单。问题是在前6行之后,弹出菜单出现在其他位置.....?如何正确显示?
我正在使用Cocoapod(SwiftPopMenu)显示弹出菜单。
@IBAction func MenuButton(_ sender: UIButton)
{
print(VideoName_For_ContextMenu)
let buttonPostion = sender.convert(sender.frame.origin, to: tableview)
showMenu(x:buttonPostion.x, y:buttonPostion.y, VideoName: VideoName_For_ContextMenu)
}.
func showMenu(x:CGFloat,y:CGFloat,VideoName:String)
{
popMenu = SwiftPopMenu(frame: CGRect(x: KSCREEN_WIDTH-185, y: y+200, width: 180, height: 112))
popMenu.popData = [(icon:"TrashBin",title:"Delete"),
(icon:"Bookmark",title:"Add BookMarks")]
popMenu.didSelectMenuBlock =
{
[weak self](index:Int)->Void in
self?.popMenu.dismiss()
print("block select \(index)")
if index == 0
{
self!.DeleteMyVideos(VideoNames: VideoName)
DispatchQueue.main.async
{
let alert = UIAlertController(title: "Success", message: "Your Video has been Deleted.....!", preferredStyle: .alert)
alert.addAction(UIAlertAction(title: "OK", style: .default, handler: nil))
self!.present(alert, animated: true, completion: nil)
self!.getData()
self!.tableview.reloadData()
}
}
}