locationOfTouch 和 numberOfTouches [英] locationOfTouch and numberOfTouches
本文介绍了locationOfTouch 和 numberOfTouches的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
嗨我有这个识别器，设置了 2 个触摸，但它只返回一个，而不是两个 CGPoint
hi I have this recognizer, set with 2 touch, but it returns only one, not two CGPoint
-(void)gestureLoad {
UIGestureRecognizer *recognizer;
recognizer = [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(numTap2:)];
[(UITapGestureRecognizer *)recognizer setNumberOfTouchesRequired:2];
[self.view addGestureRecognizer:recognizer];
self.tapRecognizer = (UITapGestureRecognizer *)recognizer;
recognizer.delegate = self;
[recognizer release];
}
- (void)numTap2:(UITapGestureRecognizer *)recognizer {
CGPoint location = [recognizer locationInView:self.view];
NSLog(@"x %f y %f",location.x, location.y);
}
据我所知，我用这两种方法循环触摸次数，但我还没有想出如何:
as I understand, I cycle the number of touch with these two methods, but I have not figured out how to:
-(CGPoint)locationOfTouch:(NSUInteger)touchIndex inView:(UIView *)view {
}
-(NSUInteger)numberOfTouches {
}
非常感谢！
推荐答案
在 numTap2 中，使用:
In numTap2, use:
CGPoint location = [recognizer locationOfTouch:touchIndex inView:self.view];
其中 touchIndex 是 0 或 1.
where touchIndex is either 0 or 1.
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