Reputation: 1
I'm trying to have the ability to remove the subview the user chooses with three taps. The problem I'm having is I can only get it to remove in the order the subviews were created. I have played around with 'viewWithTag' but can't figure out how to get it to do what I want.
Can I achieve what I want with the 'tapGesture' removing the subview from the location I tapped?
I'm a noob (as the kids say), so any help is much appreciated! Thanks!
@IBAction func unwindToParent(segue:UIStoryboardSegue){
var source = segue.sourceViewController as PropViewController
var propImage = UIImage(named: name as String!)
clipView = UIImageView(image: propImage!)
clipView.frame = CGRectMake(0, 0, 200.0, 200.0)
clipView.center = CGPoint (x: view.bounds.size.width/2, y: view.bounds.size.height/2)
clipView.contentMode = UIViewContentMode.ScaleAspectFit
clipView.userInteractionEnabled = true
clipView.multipleTouchEnabled = true
clipView.layer.cornerRadius = 10.0
setTag = tagCounter
tagCounter++
clipView.tag = setTag
addPinchGestureRecognizer(clipView)
addPanGestureRecognizer(clipView)
addRotationGestureRecognizer(clipView)
addTapGestureRecognizer(clipView)
view.addSubview(clipView)
view.bringSubviewToFront(clipView)
let recognizer = UITapGestureRecognizer(target: self, action:Selector("Trash:"))
recognizer.numberOfTapsRequired = 3
recognizer.delegate = self
clipView.addGestureRecognizer(recognizer)
}
func Trash(gesture: UITapGestureRecognizer){
clipView.viewWithTag(setTag)?.removeFromSuperview()
}
Upvotes: 0
Views: 638
Reputation: 13551
Theres probably a half dozen ways I can see this being done.
The 'simplest' is probably getting the location of the tap
tap.locationInSubview
And then checking if any of the views contain that point
uiViewObject.containtsPoint()
If it contains the point remove it
Or you could add gestures to each view and, as suggested by jrturton, uses the taps view source as the view to remove.
Upvotes: 0
Reputation: 119242
The gesture recogniser has a view property - that will be the tapped view.
Never use tags for anything.
Upvotes: 2