当使用滑动手势识别器转到Swift中的另一个视图时如何执行操作

Question

Here is my problem (on Xcode 6.0.1, coding in Swift, using iOS 7.1 and 8.0) : I am going from one view to another using a Swipe Gesture Recogniser (with the Action Segue " present modally ").

Before this I was using a button " next " to go from one view to another. I created an @IBAction for this button next that was performing, let say, a simple println("test") .

@IBAction func nextPressed(sender: AnyObject) {
    println("test")
}    

If I run my app, when I press the button Next, I go to the next view and " test " appears in the consol.

Now I do the same for a Swipe Gesture Recogniser, let say I create an @IBAction for the Swipe Gesture Recogniser (so I ctrl-click on the gesture icon next to the view controller icon at the top of the view, then drag into the code).

@IBAction func swipeGestureDone(sender: AnyObject) {
    println("test")
}    

If I run my app, when I swipe on the screen, I go to the next view but the " test " does not appear in the consol. I cannot figure out how to execute a function when I use a Swipe to go from one view to another. It is apparently because there is no " Sent Events " for a Swipe Gesture Recogniser as for a Button. I don't know how to do it. Everything seems to be well connected in the Connection inspector though.

Answer 1

Set cancelsTouchesInView of the recognizer to true . It does what it says.

Answer 2

I have a similar problem with a UITapGestureRecognizer. Here are a couple of things I found as solutions. They have not worked in my case, but maybe they will fix your issue:

Try selecting the type of object a UISwipeGestureRecognizer instead of AnyObject.

@IBAction func swipeGestureDone(sender: UISwipeGestureRecognizer) {
    println("test")  
}   

Also make sure that your view recognises gestures by adding

yourView.userInteractionEnabled = true