在UIGravityBehavior将UIView移出屏幕后，如何删除它？

Question

我试图在action属性上设置一个块，但没有成功...有什么想法吗？我知道UIView动画。方法有一个完成块，但不确定UIDynamicAnimations。

编辑:添加代码

[self.animator removeAllBehaviors];
UIGravityBehavior *gravityBehaviour = [[UIGravityBehavior alloc] initWithItems:@[self.onscreen]];
gravityBehaviour.gravityDirection = CGVectorMake(0, 10);
gravityBehaviour.action = ^{
    if(self.onscreen.frame.origin.y > [UIScreen mainScreen].bounds.size.height)
       [self.onscreen removeFromSuperview];
    NSLog(@"locations is %f, height is %f", self.onscreen.frame.origin.y, [UIScreen mainScreen].bounds.size.height);
};
[self.animator addBehavior:gravityBehaviour];

UIDynamicItemBehavior *itemBehaviour = [[UIDynamicItemBehavior alloc] initWithItems:@[self.onscreen]];
[itemBehaviour addAngularVelocity:-M_PI_2 forItem:self.onscreen];

[self.animator addBehavior:itemBehaviour];

输出反映出，即使在屏幕之外，视图也会继续移动。

Answer 1

因此，我通过在if语句中再抛出一个东西来解决这个问题: animator removeAllBehaviors；这似乎起到了作用。

Answer 2

let dynamicBehaviour = UIDynamicItemBehavior(items: [randomWord])
weak var weakBehaviour = dynamicBehaviour
weak var weakSelf = self


dynamicBehaviour.action = {

  if let currentY = weakBehaviour?.linearVelocityForItem(rndWord).y {

    if currentY > CGRectGetMaxY(self.view.frame) / 2 {
      weakSelf?.animator?.removeAllBehaviors()
      weakSelf?.randomWord?.removeFromSuperview()
    }
  }
}