UITapGestureRecognizer状态

Question

UIGestureRecognizer引用中说：“离散事件，如点击或滑动不能报告手势中的更改。”

如果手指只是触摸屏幕(而且还没有离开屏幕)，然后当手指离开屏幕时，我如何得到通知呢？

谢谢你的帮忙!

Answer 1

您可以通过使用

ObjC回答：

- (void)touchesBegan:(NSSet *)touches withEvent:(UIEvent *)event
- (void)touchesEnded:(NSSet *)touches withEvent:(UIEvent *)event

SWIFT3.0回答：

func touchesBegan(_ touches: Set<UITouch>, with event: UIEvent?)
func touchesEnded(_ touches: Set<UITouch>, with event: UIEvent?)

Answer 2

好吧，你的答案在于你的题目，即UIGestureRecognizerState。您需要在目标方法中检查gestureRecognizer的状态。像这样：

- (void)handleTap:(UITapGestureRecognizer *)sender {
           if (sender.state == UIGestureRecognizerStateEnded)     {
                // handling code     
           } 
           else {
                //do anything else
           }

}

以下是可能出现的情况：

UIGestureRecognizerStatePossible,
UIGestureRecognizerStateBegan,
UIGestureRecognizerStateChanged,
UIGestureRecognizerStateEnded,
UIGestureRecognizerStateCancelled,
UIGestureRecognizerStateFailed, 
UIGestureRecognizerStateRecognized = UIGestureRecognizerStateEnded

也许能帮上忙。