locationOfTouch和numberOfTouches

喜我有這個識別器，用2觸摸設置，但它只返回一個，不是兩個CGPointlocationOfTouch和numberOfTouches

-(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); 

}

按照我的理解，我循環觸摸的這兩種方法的數量，但我還沒有想出如何：

-(CGPoint)locationOfTouch:(NSUInteger)touchIndex inView:(UIView *)view { 

} 

-(NSUInteger)numberOfTouches { 

}

非常感謝！

來源

2011-05-12 Vins

在numTap2，使用方法：

CGPoint location = [recognizer locationOfTouch:touchIndex inView:self.view];

其中touchIndex是0或1

來源

2011-05-12 14:23:58

感謝，但我在哪裏有插入此？ – Vins 2011-05-12 14:30:52

好的完美，我開始理解，而不是你放入 - （CGPoint）locationOfTouch：（NSUInteger）touchIndex inView：（UIView *）view？ – Vins 2011-05-12 14:47:03

只是刪除這兩條消息 - 你不需要它們 – 2011-05-12 18:16:07

locationOfTouch和numberOfTouches

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