如何将TouchUpInside事件传递给UIButton，而不将它传递给父视图？

Question

我有一个启用分页的UIScrollview。这个滚动视图内有3个视图（页面）。在滚动视图的父视图中有一个轻击手势，在顶部显示并隐藏导航栏。

问题： 在我想添加按钮的页面之一中。但问题是，每当我点击这些按钮时，显示/隐藏导航栏方法也被触发。将触摸仅传递给这些按钮而不是滚动视图的父视图的最佳方式是什么？

Answer 1

NJones是在正确的轨道上，但我认为他的答案有一些问题。

我假设你想通过触摸任何按钮在你的滚动视图。在你的手势识别器的代表中，执行如下gestureRecognizer:shouldReceiveTouch:：

- (BOOL)gestureRecognizer:(UIGestureRecognizer *)recognizer shouldReceiveTouch:(UITouch *)touch { 

    UIView *gestureView = recognizer.view; 
    // gestureView is the view that the recognizer is attached to - should be the scroll view 

    CGPoint point = [touch locationInView:gestureView]; 
    UIView *touchedView = [gestureView hitTest:point withEvent:nil]; 
    // touchedView is the deepest descendant of gestureView that contains point 

    // Block the recognizer if touchedView is a UIButton, or a descendant of a UIButton 
    while (touchedView && touchedView != gestureView) { 
     if ([touchedView isKindOfClass:[UIButton class]]) 
      return NO; 
     touchedView = touchedView.superview; 
    } 
    return YES; 
}