Objectivec 如何从 UIGestureRecognizer 获取 UITouch 位置

发布于05月18日

我想从UIGestureRecognizer获取我的点击的UITouch位置，但我无法通过查看文档和其他SO问题来确定该如何操作.你们谁能给我指路吗？

- (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];
}

FIXED CODE HERE - THIS ALSO FIXES INVERTED Y AXIS

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

推荐答案

您可以在UIGestureRecognizer上使用locationInView:方法.如果对视图传递nil，此方法将返回touch 在窗口中的位置.

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

还有一个有用的委托方法gestureRecognizer:shouldReceiveTouch:.只需确保实现并将你的点击手势的委托设置为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];

倾听委托方法.

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