如何实现单指旋转手势识别器?
我知道有UIRotateGestureRecognizer已经是iOS的一部分。但是这个手势需要两个手指。如何实现只需要一个手指的类似手势识别器? AppStore中有一个游戏 - Gyrotate,它有很好的实现。任何线索都表示赞赏。谢谢。如何实现单指旋转手势识别器?
下面是代码 - 它在我的模拟器上工作。马克回答说,如果那是你正在寻找的。
// On new touch, start a new array of points
- (void) touchesBegan:(NSSet *) touches withEvent:(UIEvent *) event
{
self.points = [NSMutableArray array];
CGPoint pt = [[touches anyObject] locationInView:self];
[self.points addObject:[NSValue valueWithCGPoint:pt]];
}
// Add each point to the array
- (void) touchesMoved:(NSSet *) touches withEvent:(UIEvent *) event
{
CGPoint pt = [[touches anyObject] locationInView:self];
[self.points addObject:[NSValue valueWithCGPoint:pt]];
[self setNeedsDisplay];
}
// At the end of touches, determine whether a circle was drawn
- (void) touchesEnded:(NSSet *) touches withEvent:(UIEvent *) event
{
if (!self.points) return;
if (self.points.count < 3) return;
// Test 1: The start and end points must be between 60 pixels of each other
CGRect tcircle;
if (distance(POINT(0), POINT(self.points.count - 1)) < 60.0f)
tcircle = [self centeredRectangle];
// Test 2: Count the distance traveled in degrees. Must fall within 45 degrees of 2 PI
CGPoint center = CGPointMake(CGRectGetMidX(tcircle), CGRectGetMidY(tcircle));
float distance = ABS(acos(dotproduct(centerPoint(POINT(0), center), centerPoint(POINT(1), center))));
for (int i = 1; i < (self.points.count - 1); i++)
distance += ABS(acos(dotproduct(centerPoint(POINT(i), center), centerPoint(POINT(i+1), center))));
if ((ABS(distance - 2 * M_PI) < (M_PI/4.0f))) circle = tcircle;
[self setNeedsDisplay];
}
不是什么解决了我的问题,而是触发了导致我最终解决方案的时间。 – user292953 2010-11-03 15:51:45
分享,然后 – amok 2010-11-03 22:20:02
啊完美这就是我正在寻找,以及 2011-01-19 16:52:21
柯比特纳有一个完整的单指旋转手势识别器here。
我如何执行这个代码的旋转和翻译? – user930195 2011-11-10 14:00:55
你一直在问关于执行“两个”。如果你想检测两个单独的手势,只需使用两个(或更多)UIGestureRecogniser。 – Fattie 2014-06-05 10:18:56
我实现IQStickerView与OneFingerRotation,规模,调整大小和关闭功能。
特点: -
1)单指旋转缩放。
2)一个手指调整大小。
3)启用/取消旋转,缩放,使用属性调整大小。
4)自动管理多个IQStickerView。
5)也可以使用UIScrollView。
6)快速响应。
GitHub的复位器是在这里: - https://github.com/hackiftekhar/IQStickerView
与平移手势识别器来实现,这是使用另一个UIView的手势识别器连接到,但它应该工作与它连接到您要旋转视图。
- (void) gestureRotateButtonPan:(UIPanGestureRecognizer*) gestureRecognizer {
switch (gestureRecognizer.state) {
case UIGestureRecognizerStatePossible:
break;
case UIGestureRecognizerStateBegan:
{
CGPoint location = [gestureRecognizer translationInView:[gestureRecognizer view]];
_touchRotateStartPoint = [self convertPoint:location fromView:[gestureRecognizer view]];
}
break;
case UIGestureRecognizerStateChanged:
{
CGPoint imageLocation = CGPointMake(self.mainImageView.transform.tx + self.mainImageView.center.x
, self.mainImageView.transform.ty + self.mainImageView.center.y);
CGPoint buttonLocation = [gestureRecognizer translationInView:self];
buttonLocation.x += _touchRotateStartPoint.x;
buttonLocation.y += _touchRotateStartPoint.y;
CGFloat currentRotation = atan2(buttonLocation.y - imageLocation.y, buttonLocation.x - imageLocation.x)
- atan2(_touchRotateStartPoint.y - imageLocation.y, _touchRotateStartPoint.x - imageLocation.x);
CGFloat rotation = -(_lastRotation - currentRotation);
CGAffineTransform currentTransform = self.mainImageView.transform;
CGAffineTransform rotatedTransform = CGAffineTransformRotate(currentTransform, rotation);
self.mainImageView.transform = rotatedTransform;
[self setNeedsDisplay];
}
break;
case UIGestureRecognizerStateCancelled:
case UIGestureRecognizerStateEnded:
_lastRotation = 0.0;
break;
case UIGestureRecognizerStateFailed:
break;
default:
break;
}
}
只是FWIW - 最好的方法是从根本上理解,如果切线的导数是平滑的,那么它可能是一个很好的用户绘制的圆弧。 切线的导数非常容易计算,概念上并且只需很少的状态 - 您只需要检查最后两帧! 这是识别弧段的“魔术”算法。 – Fattie 2016-04-06 23:41:52