题
我在uiscrollview上有四个Uiviews(屏幕分为四分位数)
在四分位数上,我在每个四分位数上都有一些对象(UIImageViews)。
当用户点击屏幕时,我想找到与给定cgpoint的最接近的对象?
有任何想法吗?
我有每个四分位数中对象的cgpoint和框架(cgrect)。
更新:
(来源: Skitch.com)
红色引脚是UIImageViews。
// UIScrollView
NSLog(@" UIScrollView: %@", self);
// Here's the tap on the Window in UIScrollView's coordinates
NSLog(@"TapPoint: %3.2f, %3.2f", tapLocation.x, tapLocation.y);
// Find Distance between tap and objects
NSArray *arrayOfCGRrectObjects = [self subviews];
NSEnumerator *enumerator = [arrayOfCGRrectObjects objectEnumerator];
for (UIView *tilesOnScrollView in enumerator) {
// each tile may have 0 or more images
for ( UIView *subview in tilesOnScrollView.subviews ) {
// Is this an UIImageView?
if ( [NSStringFromClass([subview class]) isEqualToString:@"UIImageView"]) {
// Yes, here are the UIImageView details (subView)
NSLog(@"%@", subview);
// Convert CGPoint of UIImageView to CGPoint of UIScrollView for comparison...
// First, Convert CGPoint from UIScrollView to UIImageView's coordinate system for reference
CGPoint found = [subview convertPoint:tapLocation fromView:self];
NSLog(@"Converted Point from ScrollView: %3.2f, %3.2f", found.x, found.y);
// Second, Convert CGPoint from UIScrollView to Window's coordinate system for reference
found = [subview convertPoint:subview.frame.origin toView:nil];
NSLog(@"Converted Point in Window: %3.2f, %3.2f", found.x, found.y);
// Finally, use the object's CGPoint in UIScrollView's coordinates for comparison
found = [subview convertPoint:subview.frame.origin toView:self]; // self is UIScrollView (see above)
NSLog(@"Converted Point: %3.2f, %3.2f", found.x, found.y);
// Determine tap CGPoint in UIImageView's coordinate system
CGPoint localPoint = [touch locationInView:subview];
NSLog(@"LocateInView: %3.2f, %3.2f",localPoint.x, localPoint.y );
//Kalle's code
CGRect newRect = CGRectMake(found.x, found.y, 32, 39);
NSLog(@"Kalle's Distance: %3.2f",[self distanceBetweenRect:newRect andPoint:tapLocation]);
}
调试控制台
这是问题。每个瓷砖为256x256。第一个UIImageView的CGPORT转换为UISCrollView的坐标系(53.25,399.36),应在Tappoint(30,331)中死亡。为什么有区别?点击点右侧的另一个点是计算近距离(距离)?
<CALayer: 0x706a690>>
[207] TapPoint: 30.00, 331.00
[207] <UIImageView: 0x7073db0; frame = (26.624 71.68; 32 39); opaque = NO; userInteractionEnabled = NO; tag = 55; layer = <CALayer: 0x70747d0>>
[207] Converted Point from ScrollView: 3.38, 3.32
[207] Converted Point in Window: 53.25, 463.36
[207] Converted Point: 53.25, 399.36 *** Looks way off!
[207] LocateInView: 3.38, 3.32
[207] Kalle's Distance: 72.20 **** THIS IS THE TAPPED POINT
[207] <UIImageView: 0x7074fb0; frame = (41.984 43.008; 32 39); opaque = NO; userInteractionEnabled = NO; tag = 55; layer = <CALayer: 0x7074fe0>>
[207] Converted Point from ScrollView: -11.98, 31.99
[207] Converted Point in Window: 83.97, 406.02
[207] Converted Point: 83.97, 342.02
[207] LocateInView: -11.98, 31.99
207] Kalle's Distance: 55.08 ***** BUT THIS ONE's CLOSER??????
解决方案
以下方法应该解决问题。如果您发现任何怪异的东西,它可以随意指出。
- (CGFloat)distanceBetweenRect:(CGRect)rect andPoint:(CGPoint)point
{
// first of all, we check if point is inside rect. If it is, distance is zero
if (CGRectContainsPoint(rect, point)) return 0.f;
// next we see which point in rect is closest to point
CGPoint closest = rect.origin;
if (rect.origin.x + rect.size.width < point.x)
closest.x += rect.size.width; // point is far right of us
else if (point.x > rect.origin.x)
closest.x = point.x; // point above or below us
if (rect.origin.y + rect.size.height < point.y)
closest.y += rect.size.height; // point is far below us
else if (point.y > rect.origin.y)
closest.y = point.y; // point is straight left or right
// we've got a closest point; now pythagorean theorem
// distance^2 = [closest.x,y - closest.x,point.y]^2 + [closest.x,point.y - point.x,y]^2
// i.e. [closest.y-point.y]^2 + [closest.x-point.x]^2
CGFloat a = powf(closest.y-point.y, 2.f);
CGFloat b = powf(closest.x-point.x, 2.f);
return sqrtf(a + b);
}
示例输出:
CGPoint p = CGPointMake(12,12);
CGRect a = CGRectMake(5,5,10,10);
CGRect b = CGRectMake(13,11,10,10);
CGRect c = CGRectMake(50,1,10,10);
NSLog(@"distance p->a: %f", [self distanceBetweenRect:a andPoint:p]);
// 2010-08-24 13:36:39.506 app[4388:207] distance p->a: 0.000000
NSLog(@"distance p->b: %f", [self distanceBetweenRect:b andPoint:p]);
// 2010-08-24 13:38:03.149 app[4388:207] distance p->b: 1.000000
NSLog(@"distance p->c: %f", [self distanceBetweenRect:c andPoint:p]);
// 2010-08-24 13:39:52.148 app[4388:207] distance p->c: 38.013157
那里可能有更多优化的版本,因此可能值得挖掘更多。
以下方法确定两个CGPOINT之间的距离。
- (CGFloat)distanceBetweenPoint:(CGPoint)a andPoint:(CGPoint)b
{
CGFloat a2 = powf(a.x-b.x, 2.f);
CGFloat b2 = powf(a.y-b.y, 2.f);
return sqrtf(a2 + b2)
}
更新:删除fabsf(); -x^2与x^2相同,因此不必要。
更新2:添加 distanceBetweenPoint:andPoint:
方法也是完整的。
其他提示
如果您使用的是Swift,则可以计算CGPONT和CGRECT之间的距离(例如Uiview的框架)
private func distanceToRect(rect: CGRect, fromPoint point: CGPoint) -> CGFloat {
// if it's on the left then (rect.minX - point.x) > 0 and (point.x - rect.maxX) < 0
// if it's on the right then (rect.minX - point.x) < 0 and (point.x - rect.maxX) > 0
// if it's inside the rect then both of them < 0.
let dx = max(rect.minX - point.x, point.x - rect.maxX, 0)
// same as dx
let dy = max(rect.minY - point.y, point.y - rect.maxY, 0)
// if one of them == 0 then the distance is the other one.
if dx * dy == 0 {
return max(dx, dy)
} else {
// both are > 0 then the distance is the hypotenuse
return hypot(dx, dy)
}
}
谢谢@cristian,
这是您的答案的Objective-C版本
- (CGFloat)distanceToRect:(CGRect)rect fromPoint:(CGPoint)point
{
CGFloat dx = MAX(0, MAX(CGRectGetMinX(rect) - point.x, point.x - CGRectGetMaxX(rect)));
CGFloat dy = MAX(0, MAX(CGRectGetMinY(rect) - point.y, point.y - CGRectGetMaxY(rect)));
if (dx * dy == 0)
{
return MAX(dx, dy);
}
else
{
return hypot(dx, dy);
}
}
短 @Cristian 回答:
func distance(from rect: CGRect, to point: CGPoint) -> CGFloat {
let dx = max(rect.minX - point.x, point.x - rect.maxX, 0)
let dy = max(rect.minY - point.y, point.y - rect.maxY, 0)
return dx * dy == 0 ? max(dx, dy) : hypot(dx, dy)
}
就个人而言,我将其作为cgpoint扩展:
extension CGPoint {
func distance(from rect: CGRect) -> CGFloat {
let dx = max(rect.minX - x, x - rect.maxX, 0)
let dy = max(rect.minY - y, y - rect.maxY, 0)
return dx * dy == 0 ? max(dx, dy) : hypot(dx, dy)
}
}
另外,您也可以将其作为cgrect扩展名实现:
extension CGRect {
func distance(from point: CGPoint) -> CGFloat {
let dx = max(minX - point.x, point.x - maxX, 0)
let dy = max(minY - point.y, point.y - maxY, 0)
return dx * dy == 0 ? max(dx, dy) : hypot(dx, dy)
}
}
不隶属于 StackOverflow