How to set 'enabled' property for gesture recognizers in Swift 3

Question

I can't set the 'enabled' property with gesture recognizers. I tried creating recognizers both programmatically and by adding an @IBAction in Xcode and connecting it by control drag to the code.

I try to use properties like this:

swipeUpOccurred(swipeUp: UISwipeGestureRecognizer).enabled = false

But it is showing error:

Value of type '(UISwipeGestureRecognizer) ->' does not have member 'enabled'

score 5 · Accepted Answer · answered Dec 28 '16 at 12:36

5

You are trying to set the isEnabled on the action of UIGestureRecognizer. Instead of that you need to create the Outlet of UIGestureRecognizer if you want to access it inside your ViewController, so create one Outlet of that gesture with its specific type then set isEnabled property to true/false when ever you need to enable/disabled it.

@IBOutlet var swipeGest: UISwipeGestureRecognizer!

Now use isEnabled property with it.

self.swipeGest.isEnabled = false

answered Dec 28 '16 at 12:36

Nirav D

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It turned out to be that simple... Thanks! – Anton Platonov Dec 28 '16 at 12:43
@AntonPlatonov Welcome mate :) – Nirav D Dec 28 '16 at 12:43

Vladyslav Zavalykhatko · Answer 2 · 2016-12-28T12:31:33.187

0

It was renamed. Try isEnabled instead.

swipeUpOccurred(swipeUp: UISwipeGestureRecognizer).isEnabled = false

edited Dec 28 '16 at 12:31

answered Dec 28 '16 at 12:28

Vladyslav Zavalykhatko

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In this case the warning is: Value of tuple type '()' has no member 'isEnabled' – Anton Platonov Dec 28 '16 at 12:32

score 0 · Answer 3 · answered Dec 28 '16 at 12:29

0

isEnabled should be used instead

answered Dec 28 '16 at 12:29

Orest Savchak

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How to set 'enabled' property for gesture recognizers in Swift 3

3 Answers3