ios - 如何从 UIGestureRecognizer 获取 UITouch 位置

Question

我想从 UIGestureRecognizer 获取我的水龙头的 UITouch 位置，但我无法通过查看文档和其他 SO 问题来弄清楚如何。你们中的一个可以指导我吗？

- (void)handleTap:(UITapGestureRecognizer *)tapRecognizer
{
    CCLOG(@"Single tap");
    UITouch *locationOfTap = tapRecognizer; //This doesn't work

    CGPoint touchLocation = [_tileMap convertTouchToNodeSpace:locationOfTap];
    //convertTouchToNodeSpace requires UITouch

    [_cat moveToward:touchLocation];
}

此处已修复代码 - 这也修复了倒 Y 轴

CGPoint touchLocation = [[CCDirector sharedDirector] convertToGL:[self convertToNodeSpace:[tapRecognizer locationInView:[[CCDirector sharedDirector] openGLView]]]];

Answer 1

您可以在 UIGestureRecognizer 上使用 locationInView: 方法。如果为 View 传递 nil，此方法将返回窗口中触摸的位置。

- (void)handleTap:(UITapGestureRecognizer *)tapRecognizer
{
    CGPoint touchPoint = [tapRecognizer locationInView: _tileMap]
}

还有一个有用的委托(delegate)方法gestureRecognizer:shouldReceiveTouch:。只需确保实现并将您的点击手势的委托(delegate)设置为 self。

保留对手势识别器的引用。

@property UITapGestureRecognizer *theTapRecognizer;

初始化手势识别器

_theTapRecognizer = [[UITapGestureRecognizer alloc] initWithTarget: self action: @selector(someMethod:)];
_theTapRecognizer.delegate = self;
[someView addGestureRecognizer: _theTapRecognizer];

监听委托(delegate)方法。

-(BOOL)gestureRecognizer:(UIGestureRecognizer *)gestureRecognizer shouldReceiveTouch:(UITouch *)touch
{
    CGPoint touchLocation = [_tileMap convertTouchToNodeSpace: touch];
    // use your CGPoint
    return YES;
}