使用 UITouch 时如何获得触摸偏移

Question

使用UITouch 时如何获得触摸偏移？我正在做一个像遥控器这样的项目。我有一个设置为多点触控的视图，我希望视图像 Mac 的触摸板一样，所以当人们移动来控制鼠标时我需要触摸偏移。有什么想法吗？

Answer 1

这可以通过UIPanGestureRecognizer 测量从屏幕中心到当前触摸位置的对角线距离来完成。

#import <QuartzCore/QuartzCore.h>

声明手势并将其挂钩到self.view，以便整个屏幕响应触摸事件。

UIPanGestureRecognizer *pan = [[UIPanGestureRecognizer alloc] initWithTarget:self action:@selector(myPanRecognizerMethod:)];
[pan setDelegate:self];
[pan setMaximumNumberOfTouches:2];
[pan setMinimumNumberOfTouches:1];
[self.view setUserInteractionEnabled:YES];
[self.view addGestureRecognizer:pan];

然后在这个方法中，我们使用手势识别器状态：UIGestureRecognizerStateChanged 来更新和测量触摸位置和屏幕中心之间的对角线距离的整数作为触摸位置的变化。

-(void)myPanRecognizerMethod:(id)sender
{
    [[[(UITapGestureRecognizer*)sender view] layer] removeAllAnimations];
    if ([(UIPanGestureRecognizer*)sender state] == UIGestureRecognizerStateChanged) {
        CGPoint touchLocation = [sender locationOfTouch:0 inView:self.view];
        NSNumber *distanceToTouchLocation = @(sqrtf(fabsf(powf(self.view.center.x - touchLocation.x, 2) + powf(self.view.center.y - touchLocation.y, 2))));
        NSLog(@"Distance from center screen to touch location is == %@",distanceToTouchLocation);
    }
}