Question

我正在尝试使用UIPanGestureRecognizer翻转UIView。我正在使用下面的代码处理翻转视图，

- (void)handlePan:(UIPanGestureRecognizer*)gesture{

    CGPoint translation = [gesture translationInView:self.view];

    NSLog(@"PAN X VALUE %f", translation.x );

    double percentageOfWidth = translation.x / (gesture.view.frame.size.width / 2);

    float angle = (percentageOfWidth * 100) * M_PI_2 / 180.0f;

    CALayer *layer = testView.layer;
    CATransform3D flipTransform = CATransform3DIdentity;
    flipTransform.m34 = -0.002f;

    flipTransform = CATransform3DRotate(flipTransform, angle, 0.0f, 1.0f, 0.0f);
    layer.transform = flipTransform;

}

我的问题是当我平移时，有时会发生一些快速跳跃，我相信那是因为translation.x（PAN X VALUE）值从前面的几个点跳过，在我的情况下我需要它非常流畅。

非常感谢任何帮助。

提前致谢。

Answer 1

您可以使用gestureRecognizerShouldBegin方法，可以限制UIPanGestureRecognizer灵敏度。

示例：

- (BOOL)gestureRecognizerShouldBegin:(UIPanGestureRecognizer *)panGestureRecognizer {
    CGPoint translation = [panGestureRecognizer translationInView:self.view];
    return fabs(translation.y) < fabs(translation.x);
}

Answer 2

以下是我为立方体旋转解决这个问题的方法 - 只需拖动并划分：

- (void)panHandle:(UIPanGestureRecognizer*)recognizer;
{
    if ([recognizer state] == UIGestureRecognizerStateBegan)
    {
        CGPoint translation = [recognizer translationInView:[self superview]];
        _startingX = translation.x;
    }
    else if ([recognizer state] == UIGestureRecognizerStateChanged)
    {
        CGPoint translation = [recognizer translationInView:[self superview]];
        CGFloat angle = -(_startingX - translation.x) / 4;
        //Now do translation with angle
        _transitionContainer.layer.sublayerTransform = [self->_currentMetrics asTransformWithAngle:angle];
    }
    else if ([recognizer state] == UIGestureRecognizerStateEnded)
    {
        [self transitionOrReturnToOrigin];
    }
}

UIView由UIPanGestureRecognizer翻转

2 个答案: